Physics Kinematics

LOCUS LOCUS LOCUS LOCUS LOCUS 1KINEMATICS Tips on how to use the study-material effectively Read the following lines carefully to understand how you can derive the maximum benefit out of this study-material. (Don’t just read them! Follow them sincerely, and you should have an excellent chance of succeeding at the JEE). 1) Give a general reading to the study-material before coming to the class for this topic. This will ensure that you have at least an overview of what is being taught in the class and you can follow the classroom discussion in a more effective manner. 2) After the class, go through this material again, this time in detail, observing all the fine points, working out the examples on your own, completing the exercises sincerely, and pondering over the subject matter. This will complement what you have studied in the class. Particularly about the exercises, if you cannot solve any particular problem, don’t just give up and ask your teacher for the solution! Keep trying and you are sure to hit upon the solution sooner or later. 3) After going through the study-material in the detailed fashion described above, reread your classroom notes and solve on your own all the problems discussed in the class. This will ensure your complete mastery over the topic. 4) From an examination point of view, take self-tests at home on this topic so that you actually get the feel on how to solve questions on this particular topic in an exam-like situation. This will additionally help in building your examination temperament. 5) Prepare for the monthly tests seriously, as you would prepare for an actual exam. 6) Every few days, pick up this study-material and go through it again, so that the topic remains strong and fresh in your memory, and you are always exam-ready in relation to this topic. 7) Lastly, and very importantly, try to frame new questions on your own (and also solve them!), by modifying the questions discussed in this study-material. Have discussions with your friends on these ‘new’ questions. This will really help in expanding your mind’s horizons and give you hours of intellectual pleasure. w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 2 KINEMATICS REFERENCE FRAME: A reference Frame is the collection of reference axes (i.e., coordinate axes) and a reference point (i.e. origin). Every direction and distance would be defined with respect to reference directions and reference point respectively. Observations made in one frame may or may not be true for a different frame. A 2-dimensional reference frame : y x O (origin) O x, y → → reference point reference directions REST AND MOTION: The state and character of motion of any particle are not absolute quantities and actually depend on the observer. In other words we can say that the state of rest or motion of any particle are relative entities and can not be defined absolutely. To put precisely, there exists no universal, fixed or absolute frame of reference. POSITION: The position of any particle P in a given reference frame is characterized by a length (distance from origin) and a direction (angle with x or y axis, generally with x-axis). Therefore, position is a vector quantity and the vector defining the position of a particle is known as position vector r of that particle. y x O P r θ ≡ y x O P r Now, you can note that to get the position vector of any point you have to join that point to the origin. DISPLACEMENT : Suppose that a particle P moves from A to B on the path shown in the figure below, then the displacement for this duration is defined as the change in the position of the particle. Therefore, Kinematics w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 3 KINEMATICS displacement = change in position = f i r r r ∆ · − (a vector quantity) r i y x O ∆s B A r f ∆ r = r – r → → → f i Therefore, the displacement vector is obtained by joining the final position to the initial position. DISTANCE: Distance, ∆s, is defined as the actual length of the path traversed. (a scalar quantity). Here you should note that from point A to B there are many (in fact, infinite) possible paths but for all paths displacement is same y x O B ∆ r A ∆s 1 ∆s 2 ∆s 3 Note: * | | r s ∆ ≤ ∆ i.e. magnitude of displacement < distance. * Equality holds if particle moves from A to B on a straight line and it moves continuously along same direction. AVERAGE VELOCITY & AVERAGE SPEED: Average velocity is defined as the average rate of change in position w.r.t. time, i.e. average velocity, v = = f i r r r t t − ∆ · ∆ ∆ (vector) w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 4 KINEMATICS Note: * Average velocity is always along r ∆ (displacement). Average speed is defined as the average rate of covering distance i.e. average speed, v = distance covered time taken = (scalar s t ∆ ∆ Note: * | | | | r s r s v v t t ∆ ∆ ∆ ≤ ∆ ⇒ ≤ ⇒ ≤ ∆ ∆ y x O ∆s B A t ( ) 0 ∆ r v ( + ) t t 0 ∆ i.e. (magnitude of average velocity) < (average speed) INSTANTANEOUS VELOCITY : Suppose that a particle is moving in such a way that its average velocities measured for a number of different time intervals do not turn out to be constant. This particle is said to be moving with variable velocity. Then, we must seek to determine a velocity of the particle at any given instant of time, called, instantaneous velocity. It is defined as an instantaneous rate of change in position. While defining average velocity if we decrease ∆t, i.e., letting B approach A, we find that although the displacement becomes extremely small, the ratio of r ∆ & t ∆ is not necessarily a small quantity. Similarly, while growing smaller (i.e. t ∆ → 0), the displacement vector approaches a limiting direction of the tangent to the path of the particle at A. This limiting value of r t ∆ ∆ is called the y x O B A ∆ r instantaneous velocity at the point A, or the velocity of the particle at the instant t o . ∴ velocity, 0 lim t r dr v t dt ∆ → ∆ · · ∆ Note: * dr is displacement for the time interval dt * dr is always along tangent to the path v v → → dr → O y x * v is always along dr i.e. along tangent to the path. * Derivative of position w.r.t. time gives velocity. w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 5 KINEMATICS INSTANTANEOUS SPEED: We can define it exactly in the previous manner. It is defined as instantaneous rate of covering distance. Therefore, instantaneous speed (also called speed), 0 lim t s ds v t dt ∆ → ∆ · · ∆ Note: * ds is the distance travelled along the path in the time interval dt. * Displacement dr is also along the path, therefore, | | dr ds · * Derivative of distance w.r.t time gives speed. * | | | | | | dr d r ds v v v dt dt dt · ⇒ · · · i.e. magnitude of velocity = speed * If v be the speed and ˆ t be the unit vector along the tangent to the path at some moment of time, then at that moment velocity, ˆ . v v t · ACCELERATION: For a given time interval ∆t, average acceleration is defined as average rate of change in velocity, i.e., average acceleration, change in velocity time taken v a t ∆ < >· · ∆ f i f i v v t t − · − At any given instant of time t o , instantaneous acceleration (or acceleration) is defined as instantaneous rate of change in velocity, i.e. acceleration, a = 0 lim t v dv t dt ∆ → ∆ · ∆ A point traversed half a circle of radius R with constant speed u (as shown in figure). Find : (a) Duratiton of motion, t o ; (b) displacement for the time interval [0, t 0 ], assuming that at t = 0 particle is at origin; (c) average velocity for the time intervals [0, t 0 ] and [0, t 0 /2]; (d) magnitude of average velocity for the time interval [0, t 0 /2]; (e) average acceleration for the time interval [0, t 0 ]. y x O → B i O ^ j ^ speed = (constant) u Example : 1 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 6 KINEMATICS Solution: (a) As speed is constant, its average value for any interval would be same, ∴ av.speed = ⇒ 0 0 R R u t t u π π · ⇒ · (b) at t = 0 particle was at origin and at t = t 0 particle is at point B, so the vector from O to B would give the required displacement. Therefore, for [0, t 0 ] : r ∆ = ˆ 2Ri (c) for [0, t 0 ] : v = 0 ˆ 2 r Ri t t ∆ · ∆ y x O → B i O ^ j ^ ∆r = 2 ˆ R i R u π = 2 ˆ u i π for [0, t 0 /2] : v = 0 ˆ ˆ 2 r Ri Rj t t ∆ + · ∆ = 2 ˆ ˆ ( ) R i j R u π + | ` . , y x OO j ^ t 0 /2 • y R ∆ r i ^ R = ( ) 2 ˆ ˆ u i j π + (d) for [0, t 0 /2]: Magnitude of average velocity = | | v = 2 ˆ ˆ | | u i j π + = 2 2u π (e) for [0, t 0 ]: a = v t ∆ ∆ = f i f i v v t t − − = 0 0 ( ) (0) 0 v t v t − − w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 7 KINEMATICS = ˆ ˆ ( ) ( ) uj uj R u π − − | ` . , = 2 2 ˆ u j R π − y x O B O v f v i ∴ Magnitude = 2 (2 ) /( ) u R π For a moving particle, if the position vector as a function of time is given as, 2 ˆ ˆ ( ) , r t ti t j α β · − where α and β are positive constants, then find : (a) velocity as a function of time, ( ) v t ; (b) acceleration as a function of time , ( ) a t (c) & v a for the time interval [t 1 , t 2 ] (d) position of the particle when its speed is minimum. Solution: (a) ( ) v t = 2 2 ˆ ˆ ˆ ˆ ( ) ( ) ( ) dr d ti t j d ti d t j dt dt dt dt α β α β − · · − = 2 ˆ ˆ dt dt i j dt dt α β − ˆ ˆ 2 i t j α β · − (b) ( ) a t = ˆ ˆ ˆ ˆ ( 2 ) ( ) (2 ) ˆ 2 dv d i tj d i d tj j dt dt dt dt α β α β β − · · − · − (c) v = 2 1 1 1 ( ) ( ) r t r t t t − − = 2 2 2 1 1 2 2 1 ˆ ˆ ( ) ( ) ( ) t t i t t j t t α β − + − − = 2 1 ˆ ˆ ( ) i t t j α β − + a = ( is constant) a a ∵ = ˆ 2 j β − Example : 2 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 8 KINEMATICS (d) speed, v = ˆ ˆ | | | ( 2 ) | v i t j α β · − = 2 2 2 4 t α β + ∴ v min = (at 0) t α · ∴when speed is minimum, the particle is at the origin 1. For the situation given in example 1, find the magnitudes of average velocity and average acceleration for the time interval [t 0 /2, t 0 ] . 2. In 1.0 s, a particle goes from point A to point B, moving in a semicircle . The magnitude of the average velocity is: (a) 3.14 m/s (b) 2.0 m/s (c) 1.0 m/s (d) zero 1 . 0 m 3. A particle moves such that its position vector r at time t is given by: 2 2 ˆ ˆ (3 2 ) (2 4 ) r t t i t t j · + + + ; determine the magnitude and direction of its velocity and acceleration at t = 5 sec. *************** ACCELERATED MOTION: Till now we have seen that if we have the knowledge of r then we can find v by differentiating it and by differentiating v we can find the acceleration also. Now, we will proceed in the reverse order. Using a we will try to find v and using v we will try to find r . We have, dv dt = a ⇒ dv = . a dt = change in velocity in time interval dt. Therefore, the change in velocity for the time interval 0 to t is 0 0 . t t v dv a dt ∆ · · ∫ ∫ ⇒ 0 ( ) (0) . t v t v a dt − · ∫ ⇒ 0 0 ( ) (0) . . t t v t v a dt u a dt · + · + ∫ ∫ TRY YOURSELF - 1 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 9 KINEMATICS ⇒ dr dt = 0 . t u a dt + ∫ ⇒ 0 . . . t dr u dt a dt dt | ` · + . , ∫ = change in position during time interval dt. ⇒ r ∆ (displacement) for time interval [0, t] = 0 t dr ∫ = 0 0 0 . . . t t t u dt a dt dt | ` + . , ∫ ∫ ∫ ⇒ ( ) (0) r t r − = 0 0 0 0 . . . ( ) (0) . . t t t t u t a dt dt r t r ut a dt dt | ` | ` + ⇒ · + + . , . , ∫ ∫ ∫ ∫ ⇒ ( ) r t = 0 0 . . t t i r ut a dt dt | ` + + . , ∫ ∫ Note: * 0 0 0 ( ) . & ( ) . . t t t i v t u a dt r t r ut a dt dt · + · + + ∫ ∫ ∫ * If motion starts from origin, then 0 0 ( ) . . . t t r t u t a dt dt · + ∫ ∫ * If a is constant (i.e. uniform), then 2 1 & ( ) 2 i v u at r t r ut at · + · + + and, if motion starts from origin with uniform acceleration, then 2 1 ( ) 2 r t ut at · + * | | ds v dt · ⇒ | | . ds v dt · ⇒ distance, ∆s = 0 . t ds v dt · ∫ ∫ = 0 (speed). t dt ∫ * For uniform a , we have 2 1 ( ) 2 i r t r ut at · + + Here we see that ut is along u and 2 1 2 at is along a , therefore, we can assume that in the time duration ‘t’, u contributes ut and a a r (t ) u t u t = 0 r i a t 2 1 2 y x O t contributes 2 1 2 at to the displacement of the particle. w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 10 KINEMATICS Two bodies were thrown simultaneously from the same point: one, straight up, and the other, at an angle of θ = 60° to the horizontal. The initial velocity of each body is equal to u = 25 m/s. Neglecting the air drag, find the distance between the bodies t = 1.7 s later. Solution: At some time ‘t’, positions of particles are shown in figure. We have, 1 ( ) r t = 2 1 1 2 u t gt + = 2 1 ˆ ˆ 25 2 t j gt j − and 2 ( ) r t = 2 2 1 2 u t gt + r2 ( t) u 1 y x O i ^ j ^ r 2 – r 1 u 2 r 1 (t) θ g = 2 1 ˆ ˆ ( cos sin ) 2 u i u j t gt θ θ + + ∴ required distance = 2 1 r r − = ( ) 25 25 ˆ ˆ 2 3 . 2 2 i j t + − = ( ) ( ) 2 2 25 1 2 3 1.7 2 × + − × · 22 m. *************** MOTION IN ONE DIMENSION : Suppose a particle is moving along a particular straight line. In this case we need only one reference direction (parallel to line of motion) because particle will be moving either along this direction or opposite to the this direction, therefore, this single direction can define all possible motions along the given line. In this case we will drop the arrow sign from the head of the vector symbols (for example. a will be denoted as a) because any vector can be either along reference direction or opposite to reference direction. If vector is along reference direction then we define it as +ve and if it is in the opposite direction of the reference direction, then we define it as –ve. Therefore, in this case direction can be defined with the help of sign (+ve or –ve) only. Reference frame : x O +ve x direction –ve x direction Reference Point (Origin) Reference Direction Example : 3 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 11 KINEMATICS Notations: Position → x, velocity → v, speed = | v | , acceleration → a Definitions: Av.vel. change in position time taken x v t ∆ · · ∆ ; Instantaneous velocity or velocity, 0 lim t x dx v t dt ∆ → ∆ · · ∆ ; Av. acceleration, v a t ∆ · ∆ ; Instantaneous acceleration or acceleration, 0 lim t v dv a t dt ∆ → ∆ · · ∆ ; SOME CASES: X O v > 0 a < 0 x > 0 P particle X O v < 0 a < 0 x > 0 P X O v < 0 a < 0 x < 0 P X O v < 0 a > 0 x < 0 P Note: * when x is increasing: dx v dt · (rate of change of x) > 0 * when x is decreasing: dx v dt · (rate of change of x) < 0 * when v is increasing: a > 0; when v is decreasing, a < 0 ACCELERATED MOTION : According to the analysis done in the previous section, 0 0 0 , ( ) . , t t t i v u adt x t x ut adt dt · + · + + ∫ ∫ ∫ 0 0 ( ) . . t t x t ut a dt dt · + ∫ ∫ (if x i = 0). w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 12 KINEMATICS If acceleration is uniform: If the position of a moving particle, with respect to time be, 2 ( ) 1 , x t t t · + − then, find: (a) ( ), ( ) v t a t ; (b) & v a for the time interval [3, 2]; (c) displacement and distance travelled for the time interval [0, 2]. Solution: (a) ( ) 1 2 , ( ) 2 dx dv v t t a t dt dt · · − · · − (b) (3) (2) ( 5) ( 1) 4 3 2 1 x x x v t ∆ − − − − · · · · − ∆ − a = a ( ∵ a is constant) = –2 (c) displacement = (2) (0) ( 1) (1) 2 x x x ∆ · − · − − · − distance = 2 2 0 0 | | . | 1 2 | . s v dt t dt ∆ · · − ∫ ∫ = 1 2 2 1 0 2 | 1 2 | . | 1 2 | . t dt t dt − + − ∫ ∫ = 1 2 2 0 1 2 5 (1 2 ). (2 1). 2 t dt t dt − + − · ∫ ∫ ALTERNATE METHOD: (for ∆s & ∆x) : At t = 0: x = + 1 v = + 1 a = –2 Here, we see that at t = 0 particle is at x = 1 and is moving along +ve x direction, but its acceleration is along –ve x-direction, therefore, particle will move in + ve x direction with decreasing velocity and at some time (t = ½ sec) its velocity becomes zero. At the t = ½ sec the particle still has the negative acceleration, therefore, now it will start moving in the –ve x direction or we can say that it reverses its direction of motion at the t = ½ sec. Example : 4 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 13 KINEMATICS ∴ for [0, 2] : ∆x = –2 ∆s = 1/4 + 1/4 + 2 = 5 2 ∆x= –2 v t x = 0 = 1/2 = 5/4 5/4 1/4 t = 2 path of motion O –1 1 X a v t = 0 2 A particle moves in a straight line with constant acceleration. If it covers 10 m in the first second and 20 m in the next second, find its initial velocity and acceleration. Solution: Let v 0 be the initial velocity and a be the acceleration. Applying 2 0 1 2 x v t at · + for the first second of motion, we get, 2 0 0 1 10 1 1 10 2 2 a v a v · × + × × ⇒ · + ...(i) and applying the same for the first two seconds of motion, we get, 2 0 0 1 30 2 2 30 2 2 2 v a v a · × + × × ⇒ · + ...(ii) solving (i) & (ii), we get, 0 5 v · m/s and a = 10 m/s². A ball is dropped from a tower of height h. Find its speed and duration of motion when it reaches bottom of the tower. Solution: Let the ball be dropped at the instant t = 0. In this case the point from where the ball is dropped is being considered as origin and the downward direction is being considered as the +ve x direction. If v be the velocity of the ball at the bottom of the tower, then applying 2 2 2 , v u ax · + we get, 2 2 0 2 v gh · + ⇒ 2 v gh · Example : 5 Example : 6 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 14 KINEMATICS If the ball reaches the bottom at the instant t = t 0 , then applying 2 1 , 2 x ut at · + we get 2 0 1 0 2 h gt · + • t = 0 u = 0 h +ve X O g ⇒ ⇒⇒ ⇒⇒ 0 2 t h g · A stone is dropped from a balloon ascending with v 0 = 2 m/s, from a height h = 4.8 m. How much time will the ball take to reach the ground ? Solution: This case may be compared with the situation given in example 4. Here the initial velocity is in the upward direction but acceleration is in the downward direction, therefore, firstly particle will move in the upward direction with decreasing speed. After some time its velocity will become zero and then it will start moving in the downward direction. Here, I have assumed the point of projection as the origin and upward direction as the +ve x direction (as shown in the figure). Applying 2 1 , 2 x ut at · + we get, 2 0 0 1 4.8 2 9.8 2 t t − · − × ⇒ 2 0 0 9.8 4 9.6 0 t t − − · ⇒ 0 t = ( 4) 16 4 9.8 ( 9.6) 2 9.8 − − t − × × − × t =0 v 0 +ve x h g O t = t 0 ∴ t 0 = 4 16 4 9.8 9.6 [ time is not acceptable] 2 9.8 ve + + × × − × = 1.22 sec. A particle moves on a straight line with deceleration whose magnitude varies as | | a v α · , where α is a +ve constant. The initial velocity of the particle is v 0 . What distance will it traverse before it stops? What time will it take to cover that distance ? Example : 7 Example : 8 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 15 KINEMATICS Solution: We have, dv dv a v v dt dt v α α α · − ⇒ · − ⇒ · − ⇒ ⇒ 2 2 0 0 4 dx t v v t dt α α · − + ⇒ 2 2 0 0 0 0 0 0 . . 4 x t t t dx v dt v t dt t dt α α · − + ∫ ∫ ∫ ∫ v 0 a= v α 0 t x = 0 = 0 t t = 0 a v = 0 = 0 ⇒ 2 2 3 0 0 2 12 v t t x v t α α · − + ... (ii) Let particle stops at 0 , t t · then 0 ( ) 0 v t · ⇒ 0 0 0 2 t v α − · ⇒ 0 0 2 , v t α · substituting t 0 in (ii), we get x = 3 2 2 0 0 0 0 0 2 3 2 4 8 . 2 12 v v v v v α α α α α × − × + = 3 2 3 2 0 0 2 2 2 2 3 3 v v α α α α ] − + · ] ] As particle is continuously moving in same direction, required distance = 3 2 0 0 2 ( ) 3 v x t α · 1. A body starts from rest, moves in a straight line with constant acceleration and covers a distance of 64 m in 4 sec. (a) What is the final velocity? (b) How much time is required to cover half the total distance? (c) What is the distance covered in one-half the total time? (d) What is the velocity when half the total distance has been covered? (e) What is the velocity after one half the total time? TRY YOURSELF - 2 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 16 KINEMATICS 2. A particle with an initial velocity of 10m/s moves along a straight line with a constant acceleration. When the velocity of the particle is 50 m/s, the direction of its acceleration is reversed. Find the velocity of the particle when it reaches the starting point again. 3. An object is thrown vertically upward. It has a speed of 10 m/s when it has reached one half its maximum height. (a) How high does it rise? (b) What is its velocity and acceleration 1 sec after it is thrown? (c) What is its velocity and acceleration 3 s after it is thrown? (d) What is the average velocity during the first half second? 4. From an elevated point A, a stone is projected vertically upwards. When the stone reaches a distance h below A, its velocity is double of what it was at a height h above A. Show that the greatest height attained by the stone is equal to 5/3 h. 5. A point mass starts moving in a straight line with a constant acceleration α . At a time 1 t after the beginning of motion the acceleration changes sign, remaining the same in magnitude. Determine the time t from the beginning of motion in which the point mass returns to the initial position. 6. A particle of mass m moves on the x-axis as follows : it starts from rest at t = 0 from the point x = 0, and comes to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate times ( 0 < t < 1). If α denotes the instantaneous acceleration of the particle, then: (a) α cannot remain positive for all t in the interval 0 1 t ≤ ≤ (b) α cannot exceed 2 at any point in its path (c) α must be 4 ≥ at some point or points in its path (d) α must change sign during the motion, but no other assertion can be made with the information given 7. A particle moves along a straight line such that its displacement at any time t is given by 3 2 6 3 4 x t t t · − + + . What is the velocity of the particle when its acceleration is zero? 8. A motor boat starts from rest with an acceleration given by the law 2 2 / ( 4) c m s x α · + , where c is a positive constant. Given that the velocity of the boat when its displacement is 8 m is 4 m/sf find: (a) the magnitude of c (b) the position of the boat when its speed was 4.5 m/s (c) the maximum velocity of the boat. 9. An object moves along the x-axis such that is acceleration is given as a = 3 – 2 t. Find the initial speed of the object such that the particle will have the same x-coordinate at t = 5.0 s as it had at t = 0. Also find the object’s velocity at t = 5.0 s. *************** w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 17 KINEMATICS GRAPHS RATE OF CHANGE : Position-time graph for a straight line motion: (Note that if x i = 0, then ∆x = x f – x i = x f . ⇒ Position at time t≡displacement for the time interval [0, t]) θ B x 2 x 1 x t 1 t 2 t x –x 2 1 t –t 2 1 A O P x f t = ( ) From the given position-time graph it is clear that at t = t 1 the positon is x 1 and at t = t 2 the position is x 2 . If we connect A and B then the slope of the chord AB is 2 1 2 1 tan x x x t t t θ − ∆ · · − ∆ = av.velocity for the time interval [t 1 , t 2 ] ⇒ slope of the chord on x t − graph = av. velocity and slope of chord on v t − graph = v t ∆ ∆ = av. acceleration If ∆t → 0, then B approaches towards A and finally the chord joining from A to B becomes tangent to the graph at point A. We say that the chord AB becomes a tangent to the curve at A as B → A. ∴ 0 lim t ∆ → (slope of chord) = 0 lim t ∆ → (av. rate of change) ⇒ slope of the tangent = instantaneous rate of change * Slope of the tangent on x - t graph = velocity * Slope of the tangent on s - t graph = speed * Slope of the tangent on v - t graph = acceleration. x t O A B w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 18 KINEMATICS BASIC APPROACH: Go through the analysis of following graphs carefully to learn the application of concepts described on previous page. x t O x is increasing ⇒ v is +ve; slope is increasing ⇒ v is increasing ⇒ a is +ve. t O x x is increasing ⇒ v is +ve; slope is constant ⇒ v is constant ⇒ a is zero. x t O x is increasing ⇒ v is +ve; slope is decreasing ⇒ v is decreasing ⇒ a is –ve x t O x is decreasing ⇒ v is –ve slope is increasing ⇒ v is increasing ⇒ a is +ve. x t O x is decreasing ⇒ v is –ve; slope is constant ⇒ v is constant ⇒ a is zero. x is decreasing ⇒ v is –ve; slope is decreasing ⇒ v is decreasing x t O ⇒ a is –ve Example : 9 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 19 KINEMATICS I. Analyzing a graph: The graph given below describes the motion of a particle moving along the x-direction. The entire motion is being analysed in a very basic way. Carefully go through the complete analysis and see the amount of information that can be gathered from a single graph. x O t 1 t 2 t 3 t 4 t x , v >0 , a > 0 ↑ ↑ slope ↑ v x , >0 , a < 0 ↑ ↓ ↓, v v slope x , <0 , a < 0 ↓ v v slope ↓ ↓, x , <0 , a > 0 ↓ ↑ ↑, v v slope Note : ↑ ↓ denotes increasing sense and denotes decreasing sense. * At t = t 2 , slope of the tangent = 0 ⇒ at t = t 2 , v = 0 II. For the given v– t graph, plot the x – t graph, if x in = 0. For the interval (0, t 1 ): 0 v x > ⇒ is increasing; v is increasing ⇒ slope of ‘x – t’ is increasing for [t 1 , t 2 ]: v is constant and +ve ⇒ x is increasing and slope of x – t is constant. t 1 t 2 t x t v t 1 t 2 o o Here you should notice that if you draw tangents to x – t graph just before t 1 and just after t 1 then these tangents coincide with each other. I have drawn the x – t graph in this way only because velocities just before t 1 and just after t 1 are equal (in calculus you will study this property as continuity of a function), therefore, slopes of tangents on x – t graph just before t 1 and just after t 1 must be equal (In calculus you will study this property as differentiability of a function). III. For the given ‘a – v’ graph, plot ‘v – t’ graph, if v = 0 at t = 0 a O v Example : 10 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 20 KINEMATICS Solution: At t = 0. v = 0 and a > 0 ⇒ v should increase But as v increases a decreases ⇒ with increasing v, slope of ‘v – t’ must decrease. v O t *************** AREA UNDER GRAPHS : Area = dA ∫ = . v dt ∫ = ∆x = displacement t 1 t 2 t t dA v t dt = ( ). area = = . dA v dt v v t ( ) t t +d * area under v– t graph = displacement * area under | | v t − graph = | | . speed. v dt dt · ∫ ∫ = distance * area under a – t graph = . acceleration. a dt dt · ∫ ∫ = ∆v = change in velocity. * 1 2 3 0, 0, 0 A A A > < > Displacement = 1 2 3 A A A + + Distance = 1 2 3 | | | | | | A A A + + v O t A 1 A 2 A 3 For the given velocity-time graph, select the correct alternatives out of the following : (a) particle turns back exactly once during its motion (b) acceleration is always –ve (c) average acceleration for [0, t o ] is +ve (d) average velocity for [0, t 0 ] is +ve (e) average speeds for [0, t 0 ] and [t 0 , 2t 0 ] are equal t 0 2t 0 3t 0 O t V Example : 11 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 21 KINEMATICS Solution: (a) Correct : For t < t 0 , v > 0, at t = t 0 , v = 0 and for t > t 0 , v < 0 ∴ Particle’s velocity changes sign at t = t 0 . Before t = t 0 , particle is moving along the +ve x direction and after t = t 0 it is moving along –ve x direction. ∴ Particle turns back at t = t 0 . (b) Correct: Velocity is always decreasing, therefore, a < 0. Alternatively: Slope of v – t graph is always –ve, therefore acceleration is always –ve. (c) Incorrect: Slope of ‘v – t’ is constant ∴ a = constant ⇒ < a > = a ∴ for any time interval av acceleration is –ve. (d) Correct: For [0, t 0 ]: Displacement is +ve ⇒ average velocity is +ve. (e) Correct: For [0, t 0 ] and [t 0 , 2t 0 ]: M agnitudes of areas (i.e. distances) are equal, therefore average speeds are equal. ( ∵ lengths of intervals are also equal). For a particle if motion starts from the origin and its velocity varies with time according to given velocity-time graph, find: (a) the nature of acceleration (–ve or +ve); (b) the sign of average velocity for time intervals (0, t 0 ), (t 0 , 2t 0 ) and (0, 3t 0 ); (c) compare average speeds for time intervals t 0 2t 0 3t 0 t O v 0 v (0, t 0 ) and (t 0 , 2t 0 ); (d) plot acceleration-time, position-time and distance-time graphs, for the time interval (0, 2t 0 ) Solution: (a) Acceleration is always negative, because velocity is continuously decreasing Alternate: ∵ slope of v - t graph is –ve ∴ a is –ve. (b) time interval(0, t 0 ): area is +ve ⇒ displacements +ve ⇒ average velocity is +ve Example : 12 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 22 KINEMATICS (t 0 , 2t 0 ) : area is –ve ⇒ displacement is negative ⇒ av. velocity is –ve (0, 3t 0 ) : area is –ve ⇒ displacement is negative ⇒ av. velocity is –ve. (c) Magnitudes of areas for (0, t 0 ) and (t 0 , 2t 0 ) are same. Therefore, distances travelled in these two intervals (with equal lengths) are same, and hence, average speeds for these two intervals are equal to each other. (d) From the given v–t graph, we have tan 0 0 / . v t α · ∴ slope of v–t graph = tan ( ) tan θ π α · − 0 0 tan / . v t α · − · − For the time interval(0, t 0 ), velocity is +ve and is decreasing, therefore, x should increase with decreasing slope. At t = t 0 , v becomes zero, therefore, at this moment slope of x – t graph also becomes zero. After t = t o , velocity becomes negative and still decreases, therefore, x should decrease with decreasing slope. Now, I will encourage you to think about the path of this motion. Till the instant t 0 , the particle is moving along +ve x–axis; at t = t 0 , the particle momentary comes to rest and after that it starts moving in –ve x-direction. After t = t 0 , whatever distance it covers along the –ve x-direction must also be included in the net distance travelled. v t O t t v 0 O a –v 0 x O t 0 t 0 2t 0 t 0 2t 0 x t ( ) s( ) t t 0 2t 0 θ π−α = α w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 23 KINEMATICS For the given position time graph, find the time t 0 such that the velocity at t 0 is the same as average velocity for the time interval [0, t 0 ]. Example : 13 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 24 KINEMATICS Solution: It is given that av. velocity = instantaneous velocity ⇒ slope of chord = slope of tangent Therefore we should look for an instant t 0 such that the chord t = 0 to t = t 0 is also a tangent to the curve at t = t 0 . Therefore, t 0 = 50 sec. The slope of the chord OB gives the average velocity of the particle upto t = t 0 . This chord is also a tangent to the cureve at B(t = t 0 ) and hence the slope of this chord also gives the instantaneous velocity at B. w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 25 KINEMATICS 1. The figure below shows displacement-time graph of a particle. Find the time interval (starting from t = 0) during motion such that the average velocity of the particle during that period is zero. 5 t(s) x(m) 20 15 10 25 10 20 0 2. Plot the velocity-time and speed-time graphs corresponding to the position-time graph given below. x t O t 0 2t 0 3t 0 x 0 3. For the given position-time (x – t) graph find the interval in which x O t 1 t 2 t 3 t 4 t 5 t t 6 (a) velocity is zero (b) both velocity and acceleration are negative (c) velocity is positive but acceleration is negative (d) velocity is negative but acceleration is positive. TRY YOURSELF - 3 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 26 KINEMATICS 4. Plot the approximate acceleration-time, position-time and distance-time graphs for the velocity-time graph given below. v t t 0 2t 0 O 5. The velocity of a particle that moves in the positive X-direction varies with its position, as shown in figure. Find its acceleration in 2 / m s when x = 6m. v x 8m 4m 2 m/s 4 m/s 6. A particle starts moving from rest. Its acceleration (a) versus time (t) varies as shown in the figure below. The maximum speed of the particle will be: (a) 110 m/s (b) 55 m/s (c) 550 m/s (d) 660 m/s 10m/s 2 a t(s) 11 7. Repeat example 12, if the origin is chosen on ground surface (the point where the ball will hit the ground) and the upward direction is chosen as +ve X direction ************ w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 27 KINEMATICS MOTION IN A PLANE : If a particle is moving in a plane, say the X–Y plane, any two convenient mutually perpendicular directions in the x–y plane are chosen as the X and Y axes. Now, we will resolve all vectors along x and y axes. We have r = ˆ ˆ xi yj + and ˆ ˆ x y v v i v j · + But, v = dr dt ⇒ ˆ ˆ x y v i v j + = ˆ ˆ ( ) d xi yj dt + ⇒ ˆ ˆ x y v i v j + = ˆ ˆ dx dy i j dt dt + O Y X x P( , ) x y v x v v y y r ⇒ (velocity component along ) x dx v x dt · and (velocity component along ) y dy v y dt · Similarly, x a = x dv dt and y y dv a dt · Now, x, v x and a x are related by equations of straight line motion along the x-axis (the same argument holds for components along y-axis). The problem of motion in a plane is thus, broken up into two independent problems of straight line motion, one along the x-axis and the other along the y-axis. If a x and a y are constant, then 2 2 2 1 , , 2 2 x x x x x x x x v u a t x u t a t v u a x · + · + · + ; and 2 2 2 1 , , 2 2 y y y y y y y y v u a t y u t a t v u a y · + · + · + . At this juncture, I would like to encourage you to think about the sense in which these two independent sets of equations are related to each other. A particle moves in the X–Y plane, making an angle of 37° with the x-axis. At t = 0 the particle is at the origin and its velocity is 8.0 m/s along x-axis. Find the velocity and the position of the particle at t = 4.0 s. Solution: As discussed above, here we will first find the x and y components of initial velocity and acceleration. Using these components we can completely analyze the motions along these two axes. Example : 14 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 28 KINEMATICS Solution: u x = 8.0 m/s, a x = a cos 37° = (1.5 m/s²) × 4 5 u y = 0, a y = a sin 37° = (1.5 m/s²) × 3 5 = 0.9 m/s². 37° a =1.5 m/s² u = 8.0 m/s t = 0 y x At t = 4.0 s x = 2 1 1 8 4 1.2 16 41.6 . 2 2 x x u t a t m m | ` + · × + × × · . , y = 2 1 7.2 2 y y u t a t m + · x v = (8 1.2 4) 12.8 x x u a t m s m s + · + × · y v = 3.6 y y u a t m s + · a =1.5 m/s² O Y X α v y v x v y x ∴ v = 2 2 13.3 x y v v m s + · and tan α = 3.6 9 12.8 32 y x v v · · Hence, at t = 4 s, the particle is at (41.6 m, 7.2 m) and its velocity is 13.3 m/s at an angle of 1 tan (9 32) − with the +ve x-direction. Projectile motion This is the two-dimensional motion of a particle thrown obliquely into the air. It is an example of curved motion with constant acceleration. We assume that the effect of air can be neglected. O Y v X vx ( < 0) v y ( 0) vy > v Range R α g v vx = = cos v = 0 u θ ( y v x H u vy Example : 15 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 29 KINEMATICS The particle is projected with speed u (called as projection speed) at an angle α with x-axis (called as angle of projection). The motion of this particle is an accelerated one with constant acceleration ‘g’ in the downward ( –ve y) direction. The particle moves on the path shown in the figure. H is called maximum height and R is the horizontal range. Total time of motion is known as the time of flight, T. Here, I will suggest you to analyse the above the figure thoroughly and notice all the given informations carefully. We have, x u = cos , 0 x u a α · ∴ v x = cos x x u a t u α + · (constant) ....(i) and x = 2 1 2 x x u t a t + = cos . u t α ....(ii) Again, sin y u u α · , y a = –g; ∴ y v = sin y y u a t u gt α + · − ....(iii) and y = ( ) 2 2 1 1 sin 2 2 y y u t a t u t gt α + · − ...(iv) Using the four equations above we can find the velocity and position of particle at any time during its flight. Time of flight, T : At ( ) 2 1 , 0 0 sin . 2 t T y u T gT α · · ⇒ · − ⇒ 2 sin T u g α · Horizontal Range, R: At 2 sin , cos . cos . u t T x R R u T u g α α α · · ⇒ · · ⇒ 2 sin 2 u R g α · Maximum height, H: At 2 2 2 , 0 2 0 ( sin ) y y y y y H v v u a y u α · · ⇒ · + ⇒ · − ⇒ 2 2 sin 2 u H g α · Equation of trajectory: Time independent relation between x and y coordinates of any point on the path is the equation of the path. We have, x = cos . cos x u t t u α α ⇒ · w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 30 KINEMATICS again, y = 2 1 sin . 2 u t gt α − ⇒ y = 2 .sin . . cos 2 cos x g x u u u α α α | ` − . , ⇒ 2 2 2 tan 2 cos gx y x u α α · − Note: * The same approach could be followed for any type of motion in a plane. * Each point on the path must satisfy the equation of the trajectory * R(α) = R(π/2 – α). * For α ∈[0, π/2], R is symmetrical about θ = 45°. For α ∈ (0, π/4), R increases with angle and for α∈(π/4, π/2), R decreases with angle * R is maximum at θ = 45° A ball is projected horizontally with speed u from height H above ground surface (horizontal). Find the time of motion and horizontal distance travelled, before this ball hits the ground. Solution: If the ball hits the ground in T seconds, then H = 2 1 2 gT (Since ball is projected horizontally its motion in vertical direction is due to g only) ⇒ 2H T g · Again, R = u.T (As there is no horizontal acceleration, only u is responsible for horizontal movement of the particle.) = 2 . H u g ALTERNATE METHOD: x u = , 0 x x x x u a v u a t u · ⇒ · + · and x = 2 1 . 2 x x u t a t u t + · Example : 16 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 31 KINEMATICS Again, y u = 0, y y y y a g v u a t gt · + ⇒ · + · and y = 2 1 2 y y u t a t + g R H u Y x O = 2 1 2 gt If T be the time of flight, then at t = T, y = H ⇒ 2 1 2 H gT · ⇒ 2 T H g · Again, at t = T, x = R ∴ R = x(T) = u.T = 2 u H g A body projected with the same velocity at two different angles covers the same horizontal distance R. If T 1 and T 2 are the two time of flight, prove that R = 1 2 1 . 2 gT T Solution: Since it is given that the ranges of two projectiles are same, if one is projected at an angle θ, the other must be projected at 2 π θ − . Thus, T 1 = 2 sin u g θ and T 2 = 2 sin(90 ) u g θ − = 2 cos u g θ ∴ T 1 .T 2 = 2 2 2.2 sin .cos u g θ θ ∴ 1 2 g T 1 .T 2 = 2 2.2 sin .cos u g θ θ = 2 sin 2 u g θ = R Example : 17 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 32 KINEMATICS A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If α and β be the base angles and θ be the angle of projection, prove that tanθ = tan α + tan β. Solution: Let u be the projection speed and P(x, y) be the vertex of then triangle then from the figure, we have tan α = y x ... (i) and tan β = y R x − ... (ii) α θ P( , ) x y y x O x R x – u β y Again, equation of trajectory is y = x tan θ – 2 2 2 2 cos gx u θ ⇒ y = 2 tan 1 2 cos .sin gx x u θ θ θ ] − ] ] ⇒ y x = 1 x R ] − ] ] tanθ ⇒ tanθ = ( ) yR y y x R x x R x · + − − ⇒ tan θ = tan tan [using (i) & (ii)] α β + Projectile motion on Inclined plane: The figure along side shows an inclined plane at an angle β with the horizontal and a particle is projected at an angle α with the horizontal, with initial speed u. In such a case we take our reference x and y axes in the directions along and perpendicular to the incline plane plane as shown. We have, x u = cos( ) u α β − ; x a = sin g β − ; ∴ x v = cos( ) sin x x u a t u g t α β β + · − − and x = 2 2 1 1 cos( ). sin . 2 2 x x u t a t u t g t α β β + · − − u g cosβ g s i n β g Y X O β A Example : 19 Example : 18 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 33 KINEMATICS Again y u = sin( ) u α β − ; y a = cos g β − ; ∴ y v = sin( ) cos . y y u a t u g t α β β + · − − and y = 2 2 1 1 sin( ). cos . 2 2 y y u t a t u t g t α β β + · − − Now, using above equations we can analyze the motion of the particle completely. Time of flight, T : If the particle is projected at t = 0, then at t = T, y again becomes zero, i.e., at t = T, y = 0 ⇒ u sin (α − β).t – 2 1 cos . 0 2 g t β · ⇒ 2. .sin( ) cos u T g α β β − · Maximum height, H: If H be the maximum height with respect to plane, then at y = H, v y = 0 ⇒ 0 2 = 2 2. . y y u a H + ⇒ 0 = 2 [ sin( )] u α β − – ( ) 2 cos g H β ⇒ 2 2 sin ( ) 2. cos u H g α β β − · Note that the maximum height with respect to inclined plane under consideration is different from maximum height with respect to horizontal plane OA. In this case, maximum height is in fact maximum distance of particle from inclined plane. Range, R : If R be the range on inclined plane, then at t = T, x = R ⇒ R = 2 1 cos( ). sin . 2 u T g T α β β − − ⇒ 2 2 2 2 sin 2( ) 2 sin .sin ( ) cos cos u u R g g α β β α β β β − − · − Maximum Range, R max :To find maximum range on inclined plane corresponding to particular u, we use 0 dR dα · and we get, ⇒ 2 max (1 sin ) u R g β · + w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 34 KINEMATICS A child throws a ball so as to clear a wall of height h at a distance x from it. Find the minimum speed required for clearing the wall. Solution: If we join the top most point of the wall with the point of projection and consider this line as an inclined plane with inclination, 1 tan , h x β − · then distance 2 2 x h + must be maximum range on the assumed inclined plane. ∴ R max = 2 2 2 (1 sin ) u h x g β ≥ + + ⇒ u > ( ) 2 2 g h h x + + 2 2 sin h x h β ] · ] + ] ∵ ⇒ u min = ( ) 2 2 g h h x + + h β h + x 2 2 u x 1. An airplane flying horizontally at 100m/s drops a box at an elevation of 2000 m. (a) How much time is required for the box to reach earth? (b) How far does it travel horizontally while falling? (c) Find the horizontal & vertical components of its velocity when it strikes the earth. 2. A ball is thrwon from the ground into air at an angle with the horizontal. At a height of 9.1 m the velocity is observed to be v = 7.6 i + 6.1 j m/s. Find: (a) the maximum height to which it will rise. (b) horizontal distance traveled by the ball. (c) velocity of the ball at the instant of striking the ground. 3. A stone is thrown from the top of a tower of a tower of height 50 m with a velocity of 30 m per second at an angle of 0 30 above the horizontal. Find. (a) the time during which the stone will be in air (b) the distance from the tower base to where the stone will hit the ground (c) the speed with which the stone will hit the ground, (d) the angle formed by the trajectory of the stone with the horizontal at the point of hit. 4. A stone is thrown horizontally with a speed of 10 m/s from a balloon ascending witha velocity of 5 m/s and it reaches the ground in 8 sec. TRY YOURSELF - 4 Example : 20 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 35 KINEMATICS (a) How high is the balloon at the moment the stone is released? (b) How far the stone will move horizontally before striking the ground. (c) What is th impact velocity with the ground ? ( take 2 10 / g m s · ) 5. The trajectory of a projectile in a vertical plane is 2 y ax bx · − , where a,b are constants, and x and y are respectively the horizontal and vertical distances of the projectile from the point of projection. The maximum height attained is .........and the angle of projection from the horizontal is........... . 6. The maximum range of a particle with a certain speed on a horizontal plane is R. Find its maximum range when projected on an inclined plane with inclination 0 30 . 7. A gardener shower jet is placed at a distance d from the wall of a building. If R is the maximum range of the jet that is produced when the bowl is connected to the nose of a fire engine, show that the portion of the wall that is hit by the jet of water is bounded by a parabola whose height is 2 2 ( ) 2 R d R − and breadth is 2 2 2 R d − . 8. The coordinates of a bird flying in the xy-plane are 2 x t α · − and 2 y t β · , where 3.6 / m s α · and 2 1.8 / m s β · . Calculate the velocity and acceleration vectors and their magnitude as a functions of time. Also find the magnitude and direction of bird’s velocity and acceleration at t = 3.0 s. From the given data can you find whether at this instant, bird is speeding up, speeding down or it is taking a turn. If so in which direction. *********** RELATIVE MOTION : For analysing the motion of a moving body, the frame of observation can be chosen according to the convenience of the problem. The position r , velocity v and acceleration a of a particle depend upon the frame chosen. Now, we have to relate these quantities for a particle measured in two different frames. Consider two frames of reference S and S´ and suppose a particle P is observed from both the frames. The frames may be moving with respect to each other. Let, ' OO = posititon of S' w.r.t. S = ' S S r OP = posititon of P w.r.t. S = PS r w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 36 KINEMATICS ' O P = posititon of w.r.t. ' P S = ' PS r O O′ S′ X′ X S Y′ Y P we have, OP = ' OO + ' O P ⇒ ' ' PS S S PS r r r · + Differentiating both sides, we get` ' ' PS S S PS v v v · + Differentiating both sides one more time, we get ' ' PS PS S S a a a · + Note: * It is assumed that the meaning of time and d dt are same in both the frames correspondingly. * Rearranging the above equations above, we get. ' PS r = ' PS S S r r − ' PS v = ' PS S S v v − ' PS a = ' ' PS S S a a − Thus , if the velocities of two bodies are known with respect to a common frame (here S), we can find velocities of bodies with respect to each other. The same holds true for position and acceleration. * Position, velocity and acceleration of observer are always zero with respect to itself. I. Positions of points A and B are shown in some frame: • • A B r A r B x y O But if we analyze B from the frame of A, then AA r = position of A w.r.t. A 0 A A r r · − · (i.e. A would be treated as origin now) Example : 21 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 37 KINEMATICS BA B A r r r − = position of B w.r.t. A = • • A B r A O r B r B r A • ≡ A B r – r B A Note: * Vector joining B from A is position vector of B w.r.t. A. * Here, you should note that when I moved to the frame of A, I subtracted A r from both positions of A and B. If there would be a third particle C, then I would have done the same with its position vector. II. Velocities of points A and B are shown in some frame: v A v B A B B Y X O But, if we analyze B from frame of A, then AA v = velocity A of w.r.t. A = 0 A A v v − · BA B A v v v − = velocity of B w.r.t. A = Note: * Have also you should note that when I moved to the frame of A, I subtracted A v frome the velocity of each particle. In this way, we get velocities of all particles in the new frame. * We will follow the same approach for acceleration. w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 38 KINEMATICS III. At t = 0 cars A and B are moving in the same direction with constant velocities of 10 m/s and 5 m/s respectively and A behind B, find time at which A catches B if initially the distance between them is 50 m. at t = 0 at t = 0 Solution: Ground frame: 50 m B A v A =10m/s v B =5m/s at = t t 0 Let t = t 0 , be the instant at which A catches B then the distance travelled by B + 50 = distance travelled by A. ⇒ 0 0 0 50 5 50 10 10 . 5 t t t s + · ⇒ · · Let us solve for the same from the frame of B. As discussed earlier, we will subtract velocity of B from velocities of both A and B to get their velocities in the new frame. Frame of B: 50 m B A v A v B v B v B 50 m B A v B v A Rest ∴ 0 50 50 10 . 10 5 A B t s v v · · · − − It is trivial to see that B is at rest in this frame. IV. A is chasing B on a straight road. A is moving with constant speed v and B starts moving away from A with constant acceleration a when A is at a distance d away. Find the minimum value of v for B to be caught. Solution: At t = 0: A B v a d Ground frame: A B a v a a Frame of B: ≡ A B REST a v d In frame of B, if speed of A becomes zero before covering a distance d, it can never catch B. For minimum required value of v, speed of A becomes zero when it just covers the distance d, i.e., it just catches B. w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 39 KINEMATICS ∴ 0 2 = 2 min min 2 2 v ad v ad − ⇒ · Therefore, for A to be caught 2 v ad ≥ V. Two particles, 1 and 2, move with constant velocities 1 v and 2 v . At the initial moment their position vectors are 1 r and 2 r . How must these four vectors be interrelated for the particles to collide ? Solution: Ground frame: Frame of Particle 1: Analyzing motion of 2, from frame of 1, we see that 2 and 1 will collide if and only if 1 v – 2 v is antiparallel to 2 r – 1 r , i.e., v v 2 1 – r r 2 1 – 2 1 ⇒ 2 1 2 1 2 1 2 1 | | | | v v r r v v r r − − · − − − 2 1 2 1 unit vector along (unit vector along ) v v r r − ] ] · − − ] VI. Raindrops are falling on a horizontal straight road at an angle of 30° with vertical. A man is running at 10 km/h and he finds that raindrops are hitting his head vertically. Find the speed of raindrops with respect to (a) the road, (b) the moving man. Solution: Ground frame: Man’s frame: Analyzing from the moving man’s frame, we see that raindrops are falling vertically (i.e. v rm , velocity of raindrop w.r.t. moving man, is in vertically downward direction). Therefore, horizontal component of v r must cancel out v m , i.e., sin 30 r m v v ° · ⇒ 2 20 r m v v · · km/h [Ans (a)] Again, (velocity of raindrop w.r.t. moving man ) cos30 1 rm r v v · ° · km/h [Ans(b)] w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 40 KINEMATICS SWIMMER & RIVER CASE : Let us analyze different cases for a man swimming in flowing water. r v : velocity of water w.r.t. ground m v : velocity of man w.r.t. to still water mG v : velocity of man w.r.t. to ground. * r m v v : We see that, the swimmer wishes to reach point A but will reach point B. Distance AB is called drifted distance. Time of crossing, c m d t v · m v A r v r v B d mG v x drifted distance, c r r m d x t v v v · × · × * m v at an angle θ θθ θθ with r v : sin c m d t v θ · m v A r v r v B d θ m v G ≡ v m sin θ v +v cos r m θ v mG ( cos ) c r m x t v v θ · × + * Therefore, we have width velocity along width c t · velocity along flow. c x t · × * Minimum time of crossing : , sin c m d t v θ · Example : 22 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 41 KINEMATICS ∴ min c m d t v · at 2 θ π · Therefore, time of crossing is minimum when m v is along width of the river. * Zero drifting : ( . . ) net mG v i e v must be along width ∴ sin r m v v θ · ...(i) and cos net m v v θ · ...(ii) From (i), we have sin r m v v θ · . Please note that θ = 0° and θ = 90° are not possible Therefore, 0 < sinθ < 1 ⇒ 1 r m v v < ⇒ r m v v < Therefore, if river flow velocity is equal to or greater than velocity of swimmer (w.r.t. still water), it is not possible to have zero drifting at any angle. 1. An aeroplane takes off from Mumbai to Delhi with velocity 50 kph in north-east direction. Wind is blowing at 25 kph from north to south. What is the resultant displacement of aeroplane in 2 hrs. 2. Two trains, one travelling at 54 kph and the other at 72 kph, are headed towards each other on a level track. When they are two kilometers apart, both drivers simultaneously apply their brakes. If their brakes produces equal retardation in both the trains at a rate of 2 0.15 / m s , determine whether there is a collision or not. 3. A motorcycle and a car start from rest at the same place at the same time and they travel in the same direction. The cycle accelerates uniformly at 2 1 / m s up to a speed of 36 kph and the car at 2 0.5 / m s up to a speed of 54 kph. Calculate the time and the distance at which the car overtakes the cycle. 4. A river 400 m wide is flowing at a rate of 2.0 m/s. A boat is sailing at a velocity of 10.0 m/s with respect to the water, in a direction perpendicular to the river. (a) Find the time taken by the boat to reach the opposite bank. (b) How far from the point directly opposite to the starting point does the boat reach the opposite bank. 5. A swimmer wishes to cross a 500 m wide river flowing at rate 5km/hr. His speed with respect to water is 3 km/hr. (a) If he heads in a direction making an angle θ with the flow, find the time he takes to cross the river. (b) Find the shortest possible time to cross the river. TRY YOURSELF - 5 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 42 KINEMATICS 6. A man running on a horizontal road at 8 km/hr finds the rain falling vertically .He increase the speed to 12 km/hr and finds that the drops make an angle 0 30 with the vertical. Find the speed and the direction of the rain with respect to the road. 7. On a frictionless horizontal surface, assumed to be the x–y plane, a small trolley A is moving along a straight line parallel to the y-axis with a constant velocity of ( 3 1) / m s − . At a particular instant when the line OA makes an angle of 0 45 with the x-axis , a ball is thrown long the surface from the origin O. Its velocity makes an angle φ with the x-axis and it hits the trolley. 45 0 x o y A (a) The motion of the ball is observed from the frame of the trolley. Calculate the angle θ made by the velocity vector of the ball with the x-axis in this frame. (b) Find the speed of the ball with respect to the surface, if 4 / 3 φ θ · . ************ A person travelling on a straight line moves with an uniform velocity v 1 for some time and with uniform velocity v 2 for the next equal interval of time. Find the average velocity for the entire duration. Solution: If the particle moved with velocity v 1 for time t 0 , then total time of motion = 2t 0 . ∴ av. velocity = 1 0 2 0 1 2 0 displacement time 2 2 v t v t v v t + + · · A particle moves along the x-axis as x = 2 ( 2 ) ( 2 ) . u t s a t s − + − Find : (a) position of particle at t = 0 s. (b) initial velocity of particle (c) acceleration of particle. Solution: (a) x (0) = 4 (0 –2) + a (0 –2) 2 = (–2u + 4a) m. Example : 23 Example : 24 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 43 KINEMATICS (b) 2 ( 2) dx v u a t dt · · + − ∴ initial velocity = v(0) = u – 4a (c) acceleration = 2 dv a dt · Show that time taken by a particle to slide from any point on a vertical circle along a smooth chord terminating at the lowest point of the circle is constant. Solution: Consider any chord QB making an angle θ with the vertical diameter AB. For motion of particle along QB, u = 0, a = g cos θ, s = QB using s = ut + 1 2 at², we get QB = 1 2 . g cos θ. t² Q A B θ i.e. t = 2. .cos QB g θ ... (i) In right angle triangle BQA, we have cosθ = QB d where d is diameter of the given circle. substituting for cos θ in (i), we get t = 2d g = constant. If position vector of a moving particle varies with time t as (1 ), r t t ω α · − , where ω is a constant vector and α is a positive factor. Find: (a) velocity v and acceleration a of the particle as functions of time; (b) the time interval ∆t taken by particle to return to the initial point, and distance s covered during that time. Solution: Note: Initially (at t = 0) particle is at origin and r is always along constant vector ω , therefore, particle is moving along a straight line parallel to vector ω . Example : 26 Example : 25 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 44 KINEMATICS (a) (1 2 ) dr v t dt ω α · · − 2 dv a dt αω · · − (b) At t = 0 particle is at origin. Let at t = t 0 , it comes back to origin. ∴ 0 0 (1 ) 0 t t ω α − · ⇒ 0 0 (1 ) 0 t t α − · ⇒ 0 0 1 0 t or t α · · Therefore, particle comes back to origin at 1 t α · . Again, distance travelled during this time is s = 0 1 0 0 speed . | (1 2 ) | . t dt t dt α ω α · − ∫ ∫ = 1 0 | | | (1 2 ) |. t dt α ω α − ∫ = 1 2 1 0 1 2 | | (1 2 ). (2 1). t dt t dt α α α ω α α ] − + − ] ] ] ∫ ∫ = | | 2 ω α The velocity of a particle moving in the positive direction of the x axis varies as v = x α , where α is a positive constant. Assuming that at the moment t = 0 the particle was located at the point x = 0, find : (a) the time dependence of the velocity and the acceleration of the particle ; (b) the mean velocity of the particle averaged over the time that the particle takes to cover the first s meters of the path. Solution: (a) We have, v = x α Example : 27 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 45 KINEMATICS ⇒ ∴ 2 2 2 2 2 4 4 2 t d dx dt t v dt dt dt α α α | ` . , · · · · Again, 2 2 2 ( 2) . 2 2 dv d t dt a dt dt dt α α α · · · · (b) As velocity is always positive, displacement is same as distance travelled. Therefore, 2 2 ( ) ( ) . 4 t s t x t α · · Again, required average velocity = displacement time = s t = 2 2 4 2 s s s s s α α α · · The acceleration of a point started at t = 0, varies with time as shown in given graph. Find the distance travelled in 30 seconds and draw the velocity-time and position-time graphs if initial velocity and position are zero. O 10 20 30 t (second a (m/s²) 5.0 –5.0 Solution: As given in the acceleration -time graph, for time interval (0, 10) : a is constant and +ve ⇒ velocity increases linearly. ⇒ slope of x - t graph must be increasing for time interval (10, 20) : a is zero ⇒ velocity is constant ⇒ slope of x – t graph is constant Example : 28 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 46 KINEMATICS for time interval (20, 30) : a is constant and –ve ⇒ velocity decreases linearly. ⇒ slope of x - t graph is decreasing At t = 10 s, v = u + at = 0 + 5 × 10 = 50 m/s. a 0 t –5 V O t t 50 0 x 10 20 30 +5 10 20 30 10 20 30 As we see from v - t graph, velocity is always nonnegative for time interval (0, 30), therefore distance travelled, s = displacement = area under v – t graph = ( ) 1 1 10 50 50 10 10 50 2 2 | ` | ` × × + × + × × . , . , = 250 + 500 + 250 = 1000 m. Note: * At t = 30 sec, v = 0 ∴ at t = 30 sec, slope of x - t graph also becomes zero. * For time interval (0, 30) area under a - t graph is zero ⇒ change in velocity = 0 ⇒ v f in = v in = 0 m/s. w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 47 KINEMATICS A ball is released from rest from a height h above the ground surface. Assuming the point of release as origin, the downward direction as positive and gravity to be uniform, plot the following graphs for one complete trip (It is given that collision between ball and ground surface is purely elastic and duration of collision is very short): (a) velocity-time (b) speed-time (c) acceleration-time (d) displacement-time (e) position-time (f) distance-time. Solution: Let the ball be released at t = 0 so that it collides with ground at t = t 0 . Downward and upward motions of the ball are shown in following figures. h h h At 0 t= sec When (downward motion) t < t 0 When > (upward motion) t t 0 x x O x O O v g g g v • u= 0 For t < t 0 , we have 0 v u at gt gt · + · + · ; 2 2 2 1 1 1 0 2 2 2 x ut at gt gt · + · + · ; at t = t 0 x h · ⇒ h = 2 0 1 2 gt ⇒ 0 2h t g · just before collision, speed = 0 2 gt gh · just after collision, speed = 2gh For motion in upward direction (let t´ = t – t 0 , then upward motion starts at t´ = 0) initial velocity is 2gh − and acceleration is +g, therefore, using, we get v(t´) = u + at´, we get ( ') 2 ' v t gh gt · − + ⇒ 0 0 ( ) 2 ( ) v t t gh g t t − · − + − Example : 29 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 48 KINEMATICS and using x(t´) = x i + ut´ + 2 1 ' 2 at , we get 2 1 ( ') 2 ' ' 2 x t h gh t gt · − + ⇒ 2 0 0 0 1 ( ) 2 ( ) ( ) 2 x t t h gh t t g t t − · − − + − Note that at t´= t 0 (i.e. at t = 2t 0 ) v = 0 and x = 0. Graphs of x(t – t 0 ) and v(t – t 0 ) can be obtained by first plotting graphs of x(t) and v(t) and then shifting rightwards by t 0 units. a t t t t t h 2h s h O x O O 2gh speed, |v| – 2gh + 2gh v O O g t 0 2t 0 2t 0 t 0 Note: * Since motion starts from origin, displacement-time and position-time graphs are identical. * For the instant t = t 0 analysis will be done in later parts of mechanics. (This involves the concept of collisions). A car accelerates from rest at a constant rate α for sometime after which it decelerates at constant rate β to come to rest. If the total time lapse is t 0 seconds, evaluate (i) the maximum velocity reached and (ii) total distance travelled. Example : 30 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 49 KINEMATICS Solution: Let velocity is maximum at t = t 1 . For t ∈ (0, t 1 ) : a = slope of graph = tan θ 1 ⇒ α = 1 m v t ....(i) m v O v t 1 t 0 t θ 1 θ 2 For t ∈ (t 1 , t 0 ) : a = slope of graph = tan (π – θ 2 ) = –tanθ 2 0 1 0 1 ( ) m m v v t t t t β β ⇒ − · − − ⇒ · − .... (ii) Eliminating t 1 from (i) and (ii), we get. m v = 0 t αβ α β + From the graph, we see that the velocity is never negative, hence distance = displacement = area under velocity- time graph = 0 1 . . 2 m v t = 2 0 2( ) t αβ α β + For the given acceleration-time graph, plot the velocity time graph. It is given that initial velocity is v 0 and for t > t 0 , magnitude of acceleration is inversely proportional to time. t 0 t a O Example : 31 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 50 KINEMATICS Solution: From a – t graph, we see that for t ∈ (0, t 0 ) velocity must be decreasing( is ve) a – ∵ and its slope should also decrease simultaneously ( is decreasing) a ∵ in a linear fashion. t 0 t a O t 0 t O V V 0 When t ∈ (t 0 , ∞), velocity should decrease but the magnitude of its slope should be increasing. Position vector of a point relative to origin varies with time t as 2 ˆ ˆ , r ati bt j · − where a and b are positive constants. Findthe equation of the point’s trajectory. Solution: It is given taht 2 ˆ ˆ , r at i bt j · − but we know ˆ ˆ , r xi yj · + therefore, 2 and x at y bt · · − Eliminating t, we get y = 2 x b a | ` − . , ⇒ 2 2 bx y a · − For a particle moving in the x - y plane, if ˆ ˆ (3 2 ) a i j · − m/s² and ˆ ˆ (2 4 ) in v i j · + m/s, then find position of particle when its y-coordinate is maximum. Solution: we have, x a = 3 m/s², a y = –2 m/s² u x = 2 m/s, u y = 4 m/s. therefore, y v = (4 2 ), y y u a t t + · − Example : 32 Example : 33 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 51 KINEMATICS y = 2 2 1 4 , 2 y y u t a t t t + · − x v = 2 3 , x x u a t t + · + x = 2 2 1 3 2 2 2 x x u t a t t t + · + When y is maximum, 0 y v · or 0 dy dt · ⇒ 4 – 2t = 0 ⇒ t = 2 s. At t = 2 s, x = (2 × 2) + 2 3 2 2 | ` × . , = 10 m. and y = (4 × 32) – 2 (2) = 4 m. Hence, particle is at point (10, 4), and position vector of particle is r = ˆ ˆ ˆ ˆ (10 4 ) xi yj i j m + · + A particle moves in the plane xy with constant acceleration a directed along the negative y-axis. The equation of motion of the particle has the form 2 , y x x α β · − where α and β are positive constants. Find speed of particle at origin of coordinates. Solution: We have, a x = 0 and a y = – a Equation of path is y= 2 . dy dx x x dt dt α β α − ⇒ · 2 . dx x dt β − ⇒ y v = 2 . x x v x v α β − ....(i) ⇒ y dv dt = . 2 . 2 . . x x x dv dv dx x v dt dt dt α β β − − ⇒ y a = 2 . 2 . 2 . x x x a x a V α β β − − ⇒ y a = 2 2 x v β − [ ∴ a x = 0] ⇒ –a = 2 2 x v β − Example : 34 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 52 KINEMATICS ⇒ v x = 2 a β (constant) At origin: v x = 2 a β v y = . x v α [using (i) and x = 0] = . 2 a α β ∴ Speed = 2 2 2 (1 ) 2 x y a v v α β + · + If R be the range of a projectile on horizontal plane and H 1 , H 2 be maximum heights for its two possible trajectories, find the relation between the given parameters. Solution: We have, H 1 = 2 2 sin 2 u g θ ; H 2 = 2 2 2 2 sin (90 ) cos 2 2 u u g g θ θ − · ; ∴ H 1 H 2 = 2 2 2 2 2 2 2 2 2 sin . cos ( sin 2 ) 4 16 16 u u u R g g θ θ θ · · ⇒ R = 2 2 4 H H A ball is projected with some speed at an angle of 45° with the horizontal. If it just clears a wall at a distance ‘a’ from the point of projection and falls at a distance ‘b’ from the wall, then find the height of the wall. Solution: Let the projection speed be u and the projection angle be θ, then equation of path is y = 2 2 2 tan 2 cos gx x u θ θ − Example : 35 Example : 36 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 53 KINEMATICS = x.tan 45° – 2 2 2 2 cos 45 gx u × ° ( ) a+b, 0 Y ( ) a, h u a b g h θ X ⇒ y = 2 2 gx x u − Putting (a + b, 0) in the path equation, we get 0 = 2 2 2 ( ) 1 ( ) g a b g a b u u a b + + − ⇒ · + .....(i) Again, putting (a, h) in the path equation, we get, h = 2 2 2 ga a a a u a b − · − + [using (i)] ⇒ h = ab a b + A particle is projected up an inclined plane of inclination β at an elevation α to the horizontal. Show that (a) tan α = cot 2 tan , β β + if the particle strikes the plane at right angles (b) tan α = 2 tan , β if the particle strikes the plane horizontally. Solution: (a) Let t be the time of flight from O to A. Then α β u c o s ( ) α β – u u s i n ( ) α β – A X O Y t = 2 sin( ) cos u g α β β − ... (i) Now, we shall consider the motion of the particle along x-axis. Here u x = u cos (α – β); v x = 0 ( ∵ Particle strikes the inclined plane at right angles) a x = – g sin β; Applying v x = u x + a x t space along x-direction, we get 0 = cos( ) sin . u g t α β β − − Example : 37 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 54 KINEMATICS ⇒ t = cos( ) sin u g α β β − ... (ii) Dividing (i) by (ii), we get 2tan( ) cot α β β − =1 2 tan( ) cot α β β ⇒ − · ⇒ tan tan 2 1 tan . tan α β α β | ` − + . , = cot β ⇒ 2 tan 2tan α β − =cot β + tan α ⇒ tan α = cot β + 2 tan β Hence proved. (b) When the particle strikes the plane horizontally, along the vertical direction, we have, 0 = v sin α – gt ⇒ t = sin v g α ... (iii) substituting t from (iii) in (i), we get sin v g α = 2 sin( ) cos u g α β β − ⇒ sin .cos α β =2[sin . cos cos .sin ] α β α β − ⇒ 2cos .sin α β = sin .cos α β ⇒ 2tan β = tan α. Hence proved. A particle is fired at an angle θ = 60° along the direction of the breadth of a rectangular building of dimension 7m × 8m × 4m so as to sweep the edges. Find the range of the projectile. Solution: Since the projectile touches A & B, both of these points lie on the path of the projectile. After putting the coordinates of A in the trajectory equation of projectile, we obtain A B D C θ ( , ) x h ( 2 , ) x h h + 2h x x R y x u y = h = x tan θ – 2 2 2 0 2 cos gx v θ ... (i) As we know R = 2 0 sin 2 2 v x x h g θ + + · ⇒ R = 2(x + h) = 2 0 sin 2 v g θ ... (ii) Example : 38 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 55 KINEMATICS Using (i) & (ii) h = 2 2 2 tan 2 2 cos sin 2 R g h R h gR θ θ θ | ` − | ` . , − − . , ⇒ h = 2 1 tan . tan 2 2 R R h h R θ θ | ` | ` − − − . , . , ⇒ 2 2 4 cot 4 0 R Rh h θ − − · ⇒ R = 2 2 2 4 cot 16 cot 16 2 h h h θ θ t + ⇒ R = 1 cos 2 (cot cosec ) 2 sin h h θ θ θ θ + | ` + · . , ⇒ R = 2h cot (θ/2) Substituting θ = 60° & h = 4 m R = 2 × 4 cot 30° = 8 3 m. Particles are projected from the same point in a vertical plane with velocity 2gk ; prove that the locus of the vertices (points with maximum height) of their paths is ellipse : 2 4 ( ) 0. x y y k + − · Solution: If α be the angle of projection and 1 1 ( , ) x y be the coordinates of the vertex of one of the trajectories, then 1 x = 2 sin .cos 2 sin .cos u gk g g α α α α · [ ∵ 2 u gk · ] = 2k sin α. cos α and y 1 = 2 2 2 2 sin 2 sin sin 2 2 u gk k g g α α α · · Eliminating α from these two expressions, we get 2 1 x = 2 2 2 2 1 1 1 1 4 sin .cos 4 1 4 ( ) y y k k y k y k k α α | `| ` · − · − . ,. , Example : 39 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 56 KINEMATICS or 2 1 1 1 4 ( ) x y y k + − = 0 Generalising, we get 2 4 ( ) 0 x y y k + − · Hence proved. A steamer going downstream overcame a raft at a point P. 1 hr later it turned back and after some time passed the raft at a distance 6 km from the point P. Find the speed of river if speed of steamer relative to water remains constant. Solution: Let u be the speed of steamer relative to water and v be the speed of river flow. During first the 1 hr. distances travelled by the steamer and raft are (u + v) km and u km respectively. Now when the steamer turns back (say from point M), its velocity becomes (u – v) km/hr and say it passes raft at point Q after time t from the time of crossing M. Therefore, we have (u + v)×1 = (u – v)t + v (1 + t) = ut + v ⇒ t = 1 hr. Again, we have, v(1 + t) = 6 ⇒ 6 6 3 1 2 v t · · · + km/hr Alternatively: If we analyze the motion of the steamer from the raft frame, then in this frame raft and river water have zero speed and therefore time taken by steamer to come back to raft is same as time for which it had moved away. Hence, total time of motion is 2 hrs. But we know that during this interval raft has moved a distance of 6 km relative to ground. ∴ speed of raft (i.e. speed of river flow) = 6 3 2 · km/hr. Example : 40 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 57 KINEMATICS A car and a truck start moving simultaneously along the same straight line and from the same point. The car moves with constant velocity of 40 m/s and the truck starts from rest with constant acceleration of 4 m/s². Find the time t 0 that elapss before the truck catches the car. Find also the greatest distance between them prior to it and the time at which this occurs. Solution: If motion starts at t = 0, then at some time t, position of car is x 1 = 40 t and position of truck is x 2 = 2 2 1 4 2 2 t t × × · • • 4 / ² m s 40 / m s At t = 0 at t = t 1 at t = t 1 • • C T C T S x 1 x 2 When truck catches the car x 2 = x 1 ⇒ 2 0 0 0 2 40 20sec t t t · ⇒ · Before this distance between them is s = 1 2 x x − = 40 t – 2t 2 when s is maximum, 0 ds dt · ⇒ 40 – 4t = 0 ⇒ t = 10 sec and, s max = 2 (10) 40 10 2 (10) s · × − × = 400 200 200 . m − · Alternatively: Ground frame: a T T v c C Frame of truck: a T T v c a T a T • Rest ( ) v c a T C at 0 t= T C at 0 t = In this frame, as initial velocity and acceleration of car are in opposite directions and acceleration is constant, the car will first go away from the truck with decreasing speed. After some time its speed becomes zero and then it will start coming back towards the truck with increasing speed. Obviously the truck is at rest in this frame. Example : 41 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 58 KINEMATICS Let the car come back to the truck (i.e. truck catches it in ground frame) at some time t, then applying x = 2 1 2 ut at + , we get 0 = 2 2 1 1 0 40 4 2 2 c T v t a t t t − ⇒ · − × × ⇒ t = 20 sec At s max , velocity of car must be zero, therefore, applying 2 2 2 v u ax · + , we get 2 0 = 2 max max 1600 (40) 2 4 200 . 8 s s m − × × ⇒ · · If car is at maximum distance from truck at time t 0 , then applying v = u + at, we get 0 = 40 – 4 × t 0 ⇒ t 0 = 10 sec. Two ships, 1 and 2, move with constant velocities 3 m/s and 4 m/s along two mutually perpendicular straight tracks toward the intersection point O. At the moment t = 0 the ships were located at the distances 120 m and 200 m from the point O. How soon will the distancebetween the ships become the shortest and what is it equal to ? 3 m/s 120 m O 4 m/s 2 200 m Solution: Frame of ship 2: Let v 12 be the velocity of ship 1 w.r.t. ship 2. We see that in this frame ship2 is at rest and ship 1 moves along the straight line coinciding with the direction fo v 12 . The distance between the ships is minimum when ship 1 reaches the point P. Therefore, we have to find out the distance BP. 1 O 2 v 12 4ms 4ms θ 4ms 3ms ≡ 1 O B A v 12 θ Q θ P 2 (Rest) We have, v 12 (velocity of 1 relative to 2) = 2 2 4 3 5 + · m/s tanθ = 4/3 Example : 42 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 59 KINEMATICS ∴ sinθ = 4 5 and cosθ = 3 5 In triangle AQO, tanθ = QO AO ⇒ QO = AO. tanθ = 120 × 4 3 = 160 ∴ BQ = 200 – 160 = 40 m. In the same triangle, cosθ = 3 120 200 . 5 AO AQ m AQ AQ ⇒ · ⇒ · In triangle BPQ, cosθ = .cos BP BP BQ BQ θ ⇒ · ∴ Shortest distance = BP = BQ. cosθ = 3 40 24 . 5 m × · Again, sinθ = .sin QP QP BQ BQ θ ⇒ · = 4 40 32 . 5 m × · ∴ Time taken by ship 1 to reach point P is 12 AP v = 12 (200 32) 232 sec. 5 5 AQ QP m v m s + + · · = 46.4 sec. ALTERNATE METHOD: suppose that ship 1 starts moving from point B (at t = 0) and ship 2 from point A (at t = 0 only), then after time ‘t’, 1 must have travelled 4t units along +ve x-axis and 2 must have travelled 3t units along–ve y axis. Therefore, at some ‘t’ ship 1 is at point (–(200–4t),0) and ship 2 is at point (0,120–3t). Therefore, distance between 1 and 2 is ( ) ( ) 2 2 1 2 1 2 d x x y y · − + − = • (–(200–4 ), 0) t 4t 1 2 • (0, 120 – 3 ) t 3t O A B When d is minimum, d² is also minimum. Therefore, derivative of d² w.r.t. time must be zero. w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 60 KINEMATICS ∴ 2(200 4 )( 4) .(120 3 )( 3) 0 t t − − + − − · ⇒ 1600 32 720 18 0 t t − + − + · ⇒ 50t = 2320 ⇒ t = 2320/50 = 46.4 sec. Substituting t = 46.4 sec in equation (i), we will get the minimum distance, which comes out to be 24 m. An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 6 ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator. (g = 32 ft/s²) Solution: Method 1: Ground frame : Initial velocities of both elevator and coin are zero. g a 6 ft gt² 2 at² 2 Suppose that the coin falls on floor in t seconds, then the distance travelled by coin = distance travelled by floor + initial distance between coin and floor 2 1 2 gt = 2 1 6 2 at + Substituting t = 1, we get g = 12 + a ⇒ a = 32 – 12 = 20 ft/s². Method 2: Elevator frame: ≡ In this frame, elevator is at rest and coin starts falling with zero initial speed and constant acceleration (g – a). If coin falls on the floor after t seconds, then Example : 43 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 61 KINEMATICS 6 = 2 1 ( ) 2 g a t − substituting t = 1, we get a = 12 32 12 20 / ² g ft s − · − · Three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side d at t = 0. Each of the particles moves with constant speed v. A always has its velocity along AB, B along BC and C along CA. At what time will the particles meet each other ? Solution: The paths traced out by the particles A, B and C are roughly shown in the figure. By symmetry they always maintain an equilateral triangle (of reducing size) and will finally meet at the centroid of the triangle. If we focus on only two particles and find time of collision then at that instant third must be colliding with these two. Hence, we focus on motion of A A B C from the frame of B. Frame of B: We see that relative velocity of A along AB is 3 2 v . Therefore, time of collision = 2 3 3 2 d d v v · | ` . , Now, I would like to suggest you to compare this case with illustration V of example 21. Why am I asking you to do this? Because in this case velocity of A is not along B (in the frame of B) but collision still occurs. You have to figure out the reason behind this. An aeroplane has to go from a point A to another point B, 500 km away due 30° east of north. A wind is blowing due north at a speed of 20 m/s. The speed of plane with respect to air is 150 m/s. (a) Find the direction in which the pilot should head the plane to reach the point B. (b) Find the time taken by the plane to go from A to B. Example : 46 Example : 44 Example : 45 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 62 KINEMATICS Solution: Suppose that the pilot heads the plane at an angle θ with the line joining A and B (as shown in figure). As the net velocity is along the AB, we must have v w sin 30° = v P sinθ ⇒ sin θ = 1 20 1 2 150 15 × · ∴ θ = 1 1 sin 15 − | ` . , [Ans (a)] θ 30° v net v w v P East A B North Again, v net = v w cos 30° + v P cos θ = 2 2 3 15 1 20 150 2 15 | ` | ` − × + × . , . , = 10 3 10 224 + = ∴ time taken by plane to reach from A to B is ∆t = 3 1 500 10 min 60 net AB v · × × = 50 min [Ans (b)]. Two boats, A and B, move away from a buoy anchored at the middle of a river along mutually perpendicular straight lines : The boat A along the river, and the boat B across the river. Having moved off an equal distance from the buoy the boats returned. Find the ratio of the times of motion of boats, T A /T B , if the velocity of the each boat with respect to water is n = 1.2 times greater than the stream velocity. Solution: Let u be the stream velocity, then velocity of each boat(with respect to water) is v = 1.2 u. Moving away from buoy : If A moves away a distance d in time t A and B in t B , then t B = 2 2 cos . B d d d v v v u v v θ · · − = 2 2 d v u − [ ∵ v sin θ = u i.e. sin θ = u/v] θ v B v v u A = + M v u and A t = d v u + Example : 46 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 63 KINEMATICS Coming back to buoy: If A takes a time of ' A t seconds and B takes ' B t seconds in coming back to buoy, then ' B t = 2 2 d v u − and ' A t = d v u − θ v B M v v v –u A = u Therefore, total time of motion of B is B T = 2 2 2 ' 2 B B B d t t t v u + · · − and that for A is A T = ' A A d d t t v u v u + · + + − = 2 2 2vd v u − Hence B A T T = 2 2 2 2 2 2 2 2 2 2 v u d vd v u v u v v u | ` − | ` · − . , − − . , = 2 2 2 2 2 (1.2 ) (1.2) 1 1.2 1.2 u u v u v u − − − · · Hence A B T T = 2 1.2 1.8 (1.2) 1 · − A spider has fastened one end of a ‘super-elastic’ silk thread of length 1 m to a vertical wall. A small caterpillar is sitting somewhere on the thread. The hungry spider, whilst not moving from its original position starts pulling the other end of the thread with uniform speed, v 0 = 1 cm/s. Meanwhile the caterpillar starts fleeing towards the wall with a uniform speed of 1 mm/s with respect to the moving thread will the caterpillar reach the wall ? v 0 Example : 47 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 64 KINEMATICS Solution: The velocity of the thread at a distance of x meters from the wall is obviously proportionately smaller than the velocity of the end of the thread, i.e. it is xv 0 . If this value is greater than the speed of the caterpillar, then the latter will move away from the wall. Its situation will become more and more hopeless, and it will never reach the wall. x m. 1 m. v 0 On the other hand, if v caterpillar > xv 0 , the net velocity of the caterpillar is towards the wall and increases as time passes, with the consequence that the caterpillar will certainly reach the wall. The limiting case corresponds to x = v caterpillar /v 0 = 0.1 m. Starting at this point, the caterpillar does not move in either direction. A boatman sets off from one bank of a straight, uniform canal for a mark directly opposite the starting point. The speed of the water flowing in the canal is v everywhere. The boatman rows steadily at such a rate that, were there no current, the boat’s speed would also be v. He always sets the boat’s course in the direction of the mark, but the water carries him downstream. How far downstream does the water carry the boat ? What trajectory does it follow with respect to the bank? Solution: Denote the width of the canal by d and draw a straight line perpendicular to its bank at a distance d downstream from the boat’s starting point A (see figure). v net v v A F d d/2 d v The boat is initially at distance d both from F on the opposite bank and from this straight line. As both the speed of the flow and that of the boat with respect to water are v, the water flow takes the boat downstream by the same distance as is covered by the boat in the direction of F. Example : 48 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 65 KINEMATICS This means that the boat is always equally far from the point F and the straight line. The path of the boat is, therefore, a parabola with F as its focus and the straight line as its directrix. After a very long time, the boat approaches the opposite bank at a point d/2 from F. Smugglers set off in a ship in a direction perpendicular to straight shore and move at constant speed v. The costguard’s cutter is a distance a from the smugglers’ ship and leaves the shore at the same time. The cutter always move at a constant speed in the direction of the smugglers’ ship and catches up with the criminals when at a distance ‘a’ from the shore. How many times greater is the speed of costguard’s cutter than that of the smugglers’ ship ? Solution: Let kv denote the speed of coastguard’s cutter, i.e., k is the required ratio of the speeds of the two vessels. At a general time t, as shown in the figure, the distance x between the ships (initially a ) decreases by dx = kvdt – v sin α.dt in time dt. Meanwhile, the distance for the cutter from the shore increases by dy = kv sin α.dt, where α is the angle the instantaneous velocity of the cutter makes with the shore. We know, 0 0 . t v dt ∫ = a, and 0 a dx ∫ = a ⇒ 0 0 0 0 sin . t t kvdt v dt a α − · ∫ ∫ ⇒ k. a – 0 0 sin . t v dt a α · ∫ ⇒ 0 0 sin . t v dt ka a α · − ∫ Again, 0 a dy ∫ = a ⇒ 0 0 sin . t kv dt a α · ∫ ⇒ 0 0 . sin . t k v dt a α · ∫ ⇒ k.(ka – a) = a Example : 49 w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 66 KINEMATICS ⇒ 2 1 0 k k − − · ⇒ 1 5 1.68 2 k + · ≈ [This value (k≈1.68) is the famous ‘golden mean’ associated with the fibonacci series.] w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 67 KINEMATICS LEVEL – 1 1. Which statements can be possible cases in one/two dimensional motion : (a) A body has zero velocity and still be accelerating (b) The velocity of an object reverses its direction when acceleration is constant (c) An object be increasing in speed as its acceleration decreases (d) None of these. 2. A car is moving eastwards with velocity 10 m/s. In 20 sec, the velocity changes to 10 m/s northwards. The average acceleration in this time is : (a) 2 1 2 / m s towards N–W (b) 2 1 2 / m s towards N – E (c) 2 1 2 / m s towards N – W (d) 2 1 2 / m s towards N. 3. Choose the wrong statement : (a) Zero velocity of a particle does not necessarily mean that its acceleration is zero (b) Zero acceleration of a particle does not necessarily mean that its velocity is zero. (c) If speed of a particle is constant, its acceleration must be zero. (d) None of these. 4. Choose correct statements : (a) average speed is never less than magnitude of average velocity (b) average velocity is always along velocity (i.e. instantaneous velocity) (c) rate of change of speed is acceleration (d) magnitude of acceleration is rate of change of speed. 5. A small block slides without friction down an inclined plane stating from rest. Let n s be the distance travelled from t = n-1 to t = n. Then 1 n n s s + is : (a) 2 1 2 n n − (b) 2 1 2 1 n n + − (c) 2 1 2 1 n n − + (d) 2 2 1 n n + 6. Pick up the correct statements : (a) Area under a – t graph gives velocity (b) Area under a – t graph gives change in velocity (c) Path of projectile as seen by another projectile is parabola (d) A body, whatever be its motion, is always at rest in a frame of reference fixed to body itself. OBJECTIVE EXERCISE w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 68 KINEMATICS 7. A particle has initial velocity ˆ ˆ 2 4 i j + and retardation ˆ ˆ 4 8 i j + . The distance traveled by particle from t = 0 to t =1s is (a) 5 units (b) 6 units (c) 7 units (d) data is insufficient. 8. A stone projected from the ground level falls on the ground after 4 second. Then the height of the stone 1 sec after the projection is : (a) 5 m (b) 10 m (c) 15 m (d) 20 m. 9. A stone is dropped into a well in which level of water is h meters below the to top of the well. If v be the velocity of sound, the time T after which the splash is heard is : (a) 2h v (b) 2h h g v + (c) 2 2 h h g v + (d) 2 2 h h g v + . 10. A body thrown vertically up from the ground passes the height 0.2 m twice at an interval of 10 sec. What was its initial velocity. [g = 10 m/s²] (a) 52 m/s (b) 26 m/s (c) 35 m/s (d) None of these. 11. A lift starts moving from ground from rest. Its acceleration is plotted against time in the following graph. When it comes to rest its height above its starting point is : (a) 20 m (b) 64 m (c) 32 m (d) 36 m. a m s ( / ²) t s ( ) 4 0 –2 2 8 12 12. The velocity - time graph of a moving particle is shown in the figure. The maximum acceleration is (a) 1 m/s² (b) 2 m/s² (c) 3 m/s² (d) 4 m/s². 60 40 20 0 10 20 30 40 50 60 70 80 V(m/s) t s ( )→ 13. For given velocity time graph, average acceleration is zero for (a) [ ] 1 0, t (b) [ ] 1 2 , t t (c) [ ] 1 3 , t t O t 1 t 2 t 3 t 4 t 5 10 15 V (d) [ ] 2 4 , t t . w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 69 KINEMATICS 14. The co-ordinates of a moving particle at any time t are given by 2 x ct · and 2 y bt · . The speed of the particle is given by (a) 2 ( ) t c b + (b) 2 2 2t c b − (c) 2 2 t c b + (d) 2 2 2t c b + . 15. The speed at maximum height of a projectile is half of its initial velocity u. Its range on the horizontal plane is. (a) 2 2 3 u g (b) 2 3 2 u g (c) 2 3 u g (d) 2 2 u g . 16. A particle is thrown with a speed of 12 m/s at an angle 60° with the horizontal. The time interval between the moments when its speed is 10 m/s is (g 10 m/s²) : (a) 1.0 s (b) 1.2 s (c) 1.4 s (d) 1.6 s. 17. A bomber plane is moving horizontally with a speed of 500 m/s. If a bomb released from it strikes the ground in 10 sec. The angle at which it strikes the ground is (g = 10 m/s²) (a) 1 tan 5 − (b) 1 1 tan 2 − (c) 1 1 tan 5 − (d) 1 1 sin 5 − . 18. A body is projected with a speed u at an angle to the horizontal to have maximum range. At the highest point the velocity is : (a) u/2 (b) 2 u (c) u (d) 2 u . 19. Two particles are projected from same point on a horizontal plane with same initial speed v 0 in two different directions such that their horizontal ranges are same. Ratio of their maximum heights will be : (a) 2 1 tan θ (b) 0 1 sin v θ (c) 0 1 cos v θ (d) 0 1 sin v θ . 20. A particle is thrown from a platform with velocity v at an angle of 60° with the horizontal. The range obtained is R. If the platform moves horizontally in the direction of target with velocity v/2, the range will increase to : (a) 3R/2 (b) 5R/2 (c) 2R (d) 3R. 21. A boat, which has a speed of 5 km/hr in still water, crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of the river water in Km/hr is : (a) 1 (b) 3 (c) 4 (d) 41 . 22. Rain is falling vertically downwards. To a man running eastwards, the rain will appear to be coming from (a) east (b) west (c) north east (d) south west. 23. Displacement time graphs for two particles P 1 & P 2 are shown in figure. Their relative velocity : w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 70 KINEMATICS (a) is zero (b) is non zero and constant (c) continuously decreases (d) continuously increases. O P 1 P 2 x t 24. The velocity of a particle is zero at t = 0. (a) The acceleration at t = 0 must be zero (b) The acceleration at t = 0 may be zero (c) If the acceleration is zero from t = 0 to t = 10 s, the speed is zero in this interval. (d) If the speed is zero from t = 0 to t = 10 s, the acceleration is zero in this interval. w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 71 KINEMATICS LEVEL – II 1. A point moves rectilinearly. Its displacement x at time t is given by 2 2 1 x t · + . Its acceleration at time t is : (a) 3 1 x (b) 2 1 1 x x − (c) 2 t x − (d) None of these. 2. The velocity of a car moving on a straight road increases linearly according to equation V = a + bx, where a and b are positive constants. The acceleration in the course of such motion : (x is the distance travelled) (a) increases (b) decreases (c) stays constant (d) becomes zero. 3. A point moves in a straight line under the retardation Kv². If the initial velocity is u, then distance covered in ‘t’ seconds is : (a) Kut (b) ln( ) Kut K 1 (c) ln(1 ) Kut K 1 + (d) K ln(Kut). 4. Velocity of a particle moving along x-axis is given as 2 5 4 v x x · − + (in m/s) where x denotes the position of the particle in meters. The magnitude of acceleration of the particle when the velocity of the particle is zero is : (a) 0 m/s² (b) 2 m/s² (c) 3 m/s² (d) None of these. 5. The position of a particle moving along x-axis is given by x =3t² – t³ where x is in meters and t is in seconds. Consider the following statements : (i) displacement of particle after 4 s is 16 m (ii) distance traveled by particle upto 4s is 24 m (iii) displacement of the particle after 4s is –16 m (iv) distance covered by the particle upto 4s is 22 m (a) Statement (i) and (ii) only are correct (b) Statements (ii) and (iii) only are correct (c) Statements (i) and (iii) only are correct (d) none of these. 6. The velocity of a particle defined by the relation v = 8 – 0.02 x, where v is expressed in m/s and x in meter knowing that at t = 0, x = 0 then at t = 0, acceleration is (a) – 0.16 m/s² (b) – 0.02 m/s² (c) 7.08 m/s² (d) 8 m/s². 7. A particle moves according to the equation t = ax² + bx, then the acceleration of the particle when b x a · is (a) 3 a b − (b) 3 2 a g b − (c) 3 2 27 a b − (d) None of these. OBJECTIVE w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 72 KINEMATICS 8. A particle is moving with velocity 0 1 t v v T | ` · − . , . Which one is the correct graph for x - t (a) T t x O (b) T t x O (c) T t x O (d) None of these. 9. Which of the following graph correctly represents velocity-time relationship for a particle released from rest to fall freely under gravity ? (a) t V O (b) t V O (c) t V O (d) t V O . 10. A ball is dropped from a height d above the ground. It hits the ground and bounces vertically up to a height d/2. Neglecting air resistance, its velocity varies with height above the ground, as. (a) h d V O (b) h d V O (c) h d V O (d) h d V O . 11. The acceleration time graph of a particle moving on a straight line is as shown in figure. At what time the particle acquires its initial velocity ? (a) 12 sec (b) 5 sec (c) 8 sec t sec ( ) 10 4 a m/s ( ²) O (d) 16 sec. 12. The trajectory of a particle in a vertical plane is 2 6 4 y x x · − + − where x and y are respectively the horizontal and vertical distance of the projectile from the point of projection on ground. The maximum height attained is (a) 4 m (b) 5 m (c) 7 m (d) None of these. 13. A particle moves in x - y plane with a velocity v x = 8t – 2 and v y = 2. If it passes through the point x = 14 and y = 4 at t = 2 s, the equation of the path is (a) 2 2 x y y · − + (b) 2 x y · + (c) 2 2 x y · + (d) 2 2 x y y · + + . w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 73 KINEMATICS 14. A particle projected from O and moving freely under gravity strikes the horizontal plane through O at a distance R from it as shown in the figure. Then : (a) There will be two angles of projection if Rg < u² (b) the two possible angles of projection are complementary (c) The product of possible times of flight from O to A is 2 R/g (d) There will be more than two angles of projection if Rg = u². R u A α O 15. A ball is projected horizontally from an inclined plane with a velocity v 0 as shown in the figure. It will strike the plane after a time (a) 0 3 v g (b) 0 2 3 v g (c) 0 v g (d) 0 2 3v g . 60° v 0 16. Two particles start from rest simultaneously and are equally accelerated. Throughout the motion, the rela- tive velocity of one with respect to other is : (a) zero (b) none zero and is directed parallel to acceleration (c) non zero and is directed opposite to acceleration (d) non zero and is directed perpendicular to acceleration. 17. Two particles P 1 and P 2 are moving with constant velocities v 1 and v 2 respectively. Which of the statement about their relative velocity is true ? (a) it can not be greater than v 1 + v 2 (b) it can not be greater than v 1 – v 2 (c) it is always greater than v 1 + v 2 (d) it is always smaller than v 1 – v 2 . 18. What are the speeds of two objects if they move uniformly towards each other, they get 4 m closer in each second and if they move uniformly in the same direction with the original speeds they get 4 m closer in each 10 sec ? (a) 2.8 m/s and 1.2 m/s (b) 5.2 m/s and 4.6 m/s (c) 2.8 m/s and 2.1 m/s (d) 2.2 m/s and 1.8 m/s. w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 74 KINEMATICS 19. Two particles positioned at A(5, 3) and B(7, 3) are moving with constant velocity ˆ ˆ 2 3 i j + and ˆ ˆ xi yj + respectively. After 2s they collide, then the values of x and y are respectively (a) 2, 2 (b) 1, 3 (c) 3, 2 (d) 1, 1. 20. Two particles start moving simultaneously from points (0, 0) and (1, 0) respectively in the OXY plane with uniform velocities v 1 and v 2 as shown in the figure. It is found that they collide. Then (a) 1 2 2 v v · (b) 1 2 v v · (c) 2 1 2 v v · x y 45° 30° v 1 v 2 (1, 0) O (d) 2 1 3 2 v v · . 21. Pick the correct statements : (a) Average speed of a particle in given time is never less than the magnitude of the average velocity. (b) It is possible to have a situation in which ; 0 dv dt ≠ but 0 d v dt · . (c) The average velocity of a particle moving on a straight line is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval. (d) None of these. 22. Mark the correct statements for a particle moving in a straight line : (a) If the velocity & acceleration have opposite sign, the object is slowing down (b) If the position & velocity have opposite sign, the particle is moving towards the origin. (c) If the velocity is zero at an instant, the acceleration should also be zero at that instant (d) If the velocity is zero for a time interval the acceleration is zero at any instant within the time interval. w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 75 KINEMATICS SUBJECTIVE LEVEL – I 1. A person travelling on a straight line moves with a velocity v 1 for some time and with uniform velocity v 2 for the next equal time. Find average velocity. 2. A point traversed half the distance with a velocity v 0 . The remaining part of the distance was covered with velocity, v 1 for half the time, and with velocity v 2 for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion. 3. Show that time taken by a particle to slide from any point on a vertical circle along a smooth chord terminat- ing at the lowest point of the circle is constant. 4. A car accelerates from rest at a constant rate a for sometime after which it decelerates at constant rate b to come to rest. If the total time lapse is t 0 seconds, evaluate (i) the maximum velocity reached and (ii) total distance travelled. 5. A body is released from a height and falls freely towards the earth surface. Exactly 1 sec later another body is released from the same point. What is the distance between the bodies 2 sec after the release of the second body if g = 9.8 m/s² ? 6. A ball projected vertically upwards from A, the top of a tower, reaches the ground in t 1 seconds. If it is projected vertically downwards from A with the same velocity, it reaches the ground in t 2 seconds. If it falls freely from A, show that it would reach the ground in 1 2 t t seconds. 7. A ball is allowed to slip from rest down a smooth incline plane and the distances are marked every 2.0 s. If the second mark is made 1.6 m from the starting point, where are the first and fourth marks ? 8. For a particle moving in a straight line with a constant acceleration if 3 t x · + , find the position of the particle when its velocity is zero. 9. Instantaneous velocity of a particle moving in +ve x direction is given as 2 3 2 v x · + . At t = 0, particle starts from origin. Find the average velocity of the particle between the two points P(x = 2) and Q (x = 4) of its motion path. w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 76 KINEMATICS 10. For given velocity-time graph find time interval of maximum length for which av. acceleration is zero 11. For given path find point P such that av. velocity of motion from A to P is along velocity at point P. y x O A 12. Plot x – t and a – t graphs for given v – t graphs. It is given that initially particle is at x = 10. t v O t 0 13. A particle starts from the origin at t = 0. It moves in a plane with velocity given by ˆ ˆ ( cos ) v ui a t j ω ω · + . Find the equation of trajectory of the particle. 14. Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with a speed u at an angle θ with the horizontal 15. A bomb is dropped from a plane flying horizontally with uniform speed. show that the bomb will explode vertically below the plane. Is the statement true if the plane flies with uniform speed but not horizontally ? 16. If R be the range of a projectile on horizontal plane and H 1 , H 2 be the maximum heights for its two possible trajectories, find the relation between the given parameters. w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 77 KINEMATICS LEVEL – II 1. A street car moves rectilinearly from station A to the next stop B with an acceleration varying according to the law a = α – βx where α and β are positive constants and x is its distance from A. Find the distance between these two stations and maximum velocity of the car. 2. The graphs A, B and C in figure shows, the position s, velocity ds v dt · , and acceleration 2 2 dv d s a dt dt · · of a body moving along y-axis as function of time t. Which graph is for displacement, velocity and acceleration ? A B C t O x/v/a 3. For given velocity-time graph, plot approximate acceleration, speed, displacement and distance variations with respect to time t V O v 0 –v 0 2t 0 t 0 4. For given acceleration-time graph plot approximate velocity-time and position-time graphs. It is given that motion starts from rest and initially particle is at origin. a t O t 0 2t 0 SUBJECTIVE w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 78 KINEMATICS 5. For given ‘a – t’ graphs plot approximate ‘v – t’ graphs. a t O t 0 a 0 a t O t 0 2t 0 3t 0 a 0 (A) (B) 6. For given ‘v – x’ graph plot ‘x – t’ graph. v x O x 0 v 0 7. For given ‘v – x’ graph plot ‘x – t’ and ‘a – t’ graphs, if x in = +5. O x v v = x α 8. A radius vector of a point A relative to the origin (i.e. position vector) varies with time t as 2 ˆ ˆ r ti t j α β · − , where α and β are positive constants, and ˆ i and ˆ j are the unit vectors of the x and y axes. Find : (a) The equation of the point’s trajectory y(x); plot this function ; (b) velocity, speed, acceleration and magnitude of acceleration as functions of time; (c) the time dependence of the angle θ between velocity and acceleration ; (d) average velocity and its magnitude for first t seconds of motion. 9. A body is projected up such that its position vector varies with time as 2 ˆ ˆ 6 (8 5 ) r ti t t j · + − . Find the (a) initial velocity (b)initial speed (c) time of flight. 10. A cricketer hits a ball in a vertical x-y plane from the ground level with an initial velocity ˆ ˆ (10 3 30 ) u i j · + m/s. Find the time in which velocity vector makes an angle of 30° with horizontal. 11. A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 79 KINEMATICS road with an acceleration of 1 m/s² & the projection velocity in the vertical direction is 9.8 m/s. How far behind the boy will the ball fall on the car ? 12. A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8 m. Find the speed & the angle of projection (a) as seen from the truck, (b) as seen from the road. 13. A boat moves relative to water with a velocity which is 2.0 times less than the river flow velocity. At what angle to the stream direction must the boat move to minimize drifting ? 14. Two particles move in an uniform gravitational field with an acceleration (downward) g. At the initial moment the particles were located at one point and moved with velocities v 1 = 3.0 m/s and v 2 = 4.0 m/s horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular. 15. Two particles are projected simultaneously with same speed u in same vertical plane with angles of elevation θ and 2θ, where θ < 45° : At what time will their velocities be parallel ? 16. A balloon moves up vertically such that if a stone is thrown from it with a horizontal velocity v 0 relative to it, the stone always hits the ground at a fixed point 2 0 2v g horizontally away from it. Find the height of the balloon as a function of time. 17. A gun is mounted on a trolley which can move uniformly with speed v m/s along the x-axis. Two shots are fired from the origin with the gun making an angle 30° with the horizontal such that in the first case the trolley is moving along the x-axis and in the second case moving along the –ve x-axis. The respective range of the projectile is 250 m and 200 m, along the x-axis. Find the velocity of the trolley. (Assume height of the trolley to be negligible) 18. A sailor in a boat, which is going due east with a speed of 8 m/s, observes that a submarine is heading towards north at a speed of 12 m/s and sinking at a rate of 2 m/s. The commander of submarine observes a helicopter ascending at a rate of 5 m/s and heading towards west with 4 m/s. Find the actual speed of the helicopter and its speed with respect to boat. 19. Two swimmers at point A on one bank of the river have to reach point B lying right across on the other bank. One of them crosses the river along the straight line AB while the other swims at right angles to the stream and the walks the distance that he has been carried away by the stream to get to point B. What was the velocity u of his walking if both swimmers reached the destination simultaneously? The stream velocity v 0 = 2.0 km/hr and the velocity v of each swimmer with respect to water equals 2.5 km/hr. 20. What is the maximum angle to the horizontal at which a stone can be thrown and always be moving away from the thrower ? 21. An object A is kept fixed at the point x =3 m and y =1.25 m on a plank P raised above the ground. At time t = 0 the plank starts moving along the +x-direction with an acceleration 2 1.5 / m s . At the same instant a stone is projected from the origin with a velocity u as shown . w w w . m y e n g g . c o m LOCUS LOCUS LOCUS LOCUS LOCUS 80 KINEMATICS u y x O P A 1.25m 3.0 m A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of 0 45 to the horizontal. All the motions are in x-y plane. Find u and the time after which the stone hits the object. Take 2 10 / g m s · w w w . m y e n g g . c o m