Q1A train starts from rest at station A and accelerates at 0.5 m>s2 for 60 s. Afterwards it travels with a constant velocity for 15 min. It then decelerates at 1 m>s2 until it is brought to rest at station B. Determine the distance between the stations. SOLUTION Kinematics: For stage (1) motion, v0 = 0, s0 = 0, t = 60 s, and ac = 0.5 m>s2. Thus, + B 1 2 A: s = s0 + v0t + at 2 c 1 s1 = 0 + 0 + (0.5)(602) = 900 m 2 + B A: v = v0 + act v1 = 0 + 0.5(60) = 30 m>s or laws For stage (2) motion, v0 = 30 m>s, s0 = 900 m, ac = 0 and t = 15(60) = 900 s. Thus, Web) teaching in + B 1 2 A: s = s0 + v0t + copyright Wide at 2 c Dissemination permitted. World States instructors s2 = 900 + 30(900) + 0 = 27 900 m not the of is learning. on United use For stage (3) motion, v0 = 30 m>s, v = 0, s0 = 27 900 m and n ac = - 1 m>s2. Thus, and the student work for + B A: (including v = v0 + act the protected of solely work is assessing 0 = 30 + (- 1)t this work and of integrity provided t = 30 s This is the part courses 1 2 any and + s = s0 + v0t + at : of destroy 2 c their will sale 1 s3 = 27 900 + 30(30) + (- 1)(302) 2 = 28 350 m = 28.4 km Ans. 9t2 + 15t2 ft. or laws sT = 7 + 7 + 25 + (25 . Web) in copyright Wide Dissemination permitted. on United use and by the student work for (including the protected of solely work is assessing this work and of integrity provided This is the part courses any and of destroy their will sale .18 ft Ans. s = . World States instructors not the of is learning. s = .Q2 The position of a particle on a straight line is given by s = 1t3 . Determine the position of the particle when t = 6 s and the total distance it travels during the 6-s time interval.25 ft t = 6 s. where t is in seconds. s = 7 ft t = 5 s.18t + 15 dt v = 0 when t = 1 s and t = 5 s t = 0. s = 0 t = 1 s. SOLUTION s = t3 .9t2 + 15t ds v = = 3t2 .18) = 46 ft Ans. Hint: Plot the path to determine the total distance traveled. 568 s is learning. respectively. Determine the distance d to where it will land.Q3 A golf ball is struck with a velocity of 80 ft>s as shown.2)t2 (2) Web) 2 in copyright Solving Eqs.89t (1) Vertical Motion: The vertical component of initial velocity is (v0)y = 80 sin 55° = 65. + B A: sx = (s0)x + (v0)x t d cos 10° = 0 + 45. vA 80 ft/s B 45 A 10 d SOLUTION Horizontal Motion: The horizontal component of velocity is (v0)x = 80 cos 55° = 45. d = 166 ft Ans. States instructors not the of t = 3.The initial and final horizontal positions are (s0)x = 0 and sx = d cos 10°. on United use and by the student work for (including the protected of solely work is assessing this work and of integrity provided This is the part courses any and of destroy their will sale .89 ft>s.53t + (-32. respectively. (1) and (2) yields Wide Dissemination permitted. The initial and final vertical positions are (s0)y = 0 and sy = d sin 10°.53 ft>s. 1 (+ c) sy = (s0)y + (v0)y t + (a ) t2 2 cy or laws 1 d sin 10° = 0 + 65. 536)2 + (8.15(18.082 an = = = 8.453 s v = 0.3 ft s2 instructors Ans. A 300 ft SOLUTION v t B dv = 0.3t2 ft>s2.05t3 t = 18.453 s = 5. 300 ft 60° where t is in seconds.Q4 When the motorcyclist is at A.1 ft>s Ans.05t3 When s = p3 (300) ft.15t2 s t ds = 0. not the of is learning. If he starts from rest at A. or # laws at = v = 0.536 ft>s2 Web) teaching in v2 51. on United use and by the student work for (including the protected of solely work is assessing this work and of integrity provided This is the part courses any and of destroy their will sale .696 ft>s2 copyright Wide r 300 Dissemination permitted. World States a = a2t + a2n = (5.3t|t = 18.15t2 dt L0 L0 s = 0.453)2 = 51.696)2 = 10. determine the magnitudes of his velocity and acceleration when he reaches B. he increases his speed along # the vertical circular path at the rate of v = 10. p 3 (300) = 0.3tdt L0 L0 v = 0.08 ft>s = 51. 52472 = 3.57 m |d2y>dx2| |0.02x)2]3>2 r = = 2 = 190. at = vA = 3 m>s.57 # Here. where its speed is vA = 10 m>s.02| x = 60 m To determine the normal acceleration.05 m s2 teaching Ans. v2 102 an = = = 0.Q5 A toboggan is traveling down along a curve which can be y approximated by the parabola y = 0. Here.5247 m>s2 r 190. World States instructors not the of is learning. in copyright Wide Dissemination permitted. the magnitude of acceleration is or laws Web) a = a2t + a2n = 32 + 0. then dx dx [1 + (dy>dx)2]3>2 [1 + (0. 12–20.01x2 vA = 3 m>s2.02x and 2 = 0.01x2. Thus.02. = 0. Determine the magnitude of its acceleration when it reaches point A. apply Eq. on United use and by the student work for (including the protected of solely work is assessing this work and of integrity provided This is the part courses any and of destroy their will sale . A 36 m SOLUTION x 60 m Acceleration: The radius of curvature of the path at point A must be determined dy d2y first. and it is increasing at the rate of # y = 0. 2 ¢sB = .¢sC 2 ¢sB = . down 4 ft. not Wide permitted. Q6 Thus.sC + sB = l2 ¢sB = .4 2 ¢sB = ¢sC ¢sA = . Web) or A C B .¢sA 2 sA + 2 sC = l1 SOLUTION sB . copyright work on the and Dissemination in World is teaching Ans.2 ft = 2 ft c and their sale work is courses provided of will is any protected destroy and part solely assessing of the Determine the displacement of the block B if A is pulled this by for integrity United the work student use (including States of of the instructors learning. as shown in the figure below. is 4 s B + s A + L const = L total Differentiating this equation with respect to time yields the time rates of change of position. or velocities of each weight. While the physical system is somewhat more complex than the original.2 m/s.Q-7At what rate. weight A will move up at 1. and their lengths are combined and called Lconst. There are four cord segments around the four pulleys that do not change as the weights move. There are 4 equivalent-length sections that lengthen as weight B is lowered. the solution is not very different. The total cord length. so the motion of one of the weights is dependent upon one the other.3 m/s? DATUM sB sA 0. and in which direction must weight A move if weight B is to fall at a rate of 0. 2-pulley example.3 m/s B A Solution There is a single cord between weights A and B. if weight B moves down at 0. then. The total length of the cord is made up of 9 segments. The length of the cord segment connected to weight A is labeled SA. These segments are shown in red in the figure. The length of each of these sections is labeled SB. ds B ds A 4 + +0=0 dt dt 4 v B = −v A So.3 m/s. . 10.15i .69)2 + (16. B Determine the relative velocity and relative acceleration of vB car A with respect to car B at this instant.93j6 ft>s vA/B = 2( -37.1 a b = 67.1 sin 60°i .25 sin 60°j .70 not u = tan .90.1 a b = 57.93)2 = 98.Q8 Cars A and B are traveling around the circular race track. rA 300 ft 60 rB 250 ft SOLUTION vA = vB + vA>B . World States instructors 16.70j} ft>s2 in copyright Wide aA>B = 2(10.8 ft>s2 Ans.69i + 16.4 ft>s Ans. A has a speed of 90 ft>s and is increasing its speed at the rate of 15 ft>s2.44.5i .93 u = tan . learning.5)2 + ( -90. 37. 90. vA At the instant shown.5 aA = aB + aA>B 19022 . j = 25 cos 60°i . permitted.70)2 = 19.1 cos 60°j + aA>B or 300 Web) teaching aA>B = {10.6° d Ans.69 on United use and by the student work for (including the protected of solely work is assessing this work and of integrity provided This is the part courses any and of destroy their will sale .44. whereas B has a A speed of 105 ft>s and is decreasing its speed at 25 ft>s2.90i = -105 sin 30° i + 105 cos 30°j + vA>B vA>B = 5-37.4° a the of Ans. 7 not the of is aB = aA + aB>A learning.952 + (. while A maintains a constant speed. 12–35 gives World r 0.95 . cars A and B travel at speeds of 70 mi>h and 50 mi>h. is the courses any and And its direction is of destroy their 833.43 cos 30°)i + (1100 cos 30° .1 = 46.09)2 = 3737 mi>h2 Ans.3571. the magnitude of the relative velocity vB/A is yB>A = 225.09 f = tan .26.70 u = tan .1 will = 12.95i . Applying Eq.09j} mi>h2 of work is assessing Thus.43 mi>h2. The direction of the relative velocity is the same as the direction of that for relative acceleration. If B is increasing its speed by 1100 mi>h2.Car B moves along a curve having a radius of curvature of 0. its normal y2B 502 permitted. the magnitude of the relative velocity aB/A is this and of integrity provided aB>A = 23642.02 + ( .43 sin 30°)j = 0 + aB>A work (including the aB>A = {3642.26.9° c Ans.0 Web) teaching in copyright Wide Relative Acceleration: Since car B is traveling along a curve.Q9 At the instant shown. determine the velocity and acceleration of B with respect to A.9° c Ans. Thus or laws 26. 25. acceleration is (aB)n = = = 3571. on United use and by the (1100 sin 30° + 3571.833. respectively.70)2 = 36. A vA 70 mi/h vB 50 mi/h B SOLUTION 30 Relative Velocity: vB = vA + vB>A 50 sin 30°i + 50 cos 30°j = 70j + vB>A vB>A = {25.7 mi.6 mi>h Ans. sale 3642.70j} mi>h Thus.833.0i .