3/3/2014Chapter 3 Homework Chapter 3 Homework Due: 10:00pm on Friday, February 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Exercise 3.7 The coordinates of a bird flying in the xy-plane are given by and β = 1.2 m/s2 . x(t) = αt and y(t) = 3.0m − βt2 , where α = 2.4 m/s Part A Calculate the velocity vector of the bird as a function of time. Give your answer as a pair of components separated by a comma. For example, if you think the x component is 3t and the y component is 4t, then you should enter 3t,4t. Express your answer using two significant figures for all coefficients. ANSWER: v⃗ (t) = 2.4, −2.4t m/s Correct Part B Calculate the acceleration vector of the bird as a function of time. Give your answer as a pair of components separated by a comma. For example, if you think the x component is 3t and the y component is 4t, then you should enter 3t,4t. Express your answer using two significant figures for all coefficients. ANSWER: a⃗ (t) = 0,-2.4 2 m/s Correct Part C Calculate the magnitude of the bird's velocity at . t = 2.0s Express your answer using two significant figures. ANSWER: v = 5.4 m/s Correct Typesetting math: 43% http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 1/21 3/3/2014 Chapter 3 Homework Part D Let the direction be the angle, that the vector makes with the +x-axis measured counterclockwise. Calculate the direction of the bird's velocity at t = 2.0s. Express your answer in degrees using two significant figures. ANSWER: θ = -63 ∘ Correct Part E Calculate the magnitude of the bird's acceleration at . t = 2.0s Express your answer using two significant figures. ANSWER: 2.4 2 m/s Correct Part F Calculate the direction of the bird's acceleration at . t = 2.0s ANSWER: θ = -90 ∘ Correct Part G At , is the bird speeding up, slowing down or moving at constant speed? t = 2.0s ANSWER: speeding up slowing down moving at constant speed http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 2/21 3/3/2014 Chapter 3 Homework Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: 1. An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, vx , is constant. 2. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by g, is equal to 9.80 m/s2 near the surface of the earth. Hence, the y component of its velocity, vy , changes continuously. 3. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time t the projectile is in the air. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t0 = 0 s corresponds to the moment just after the ball is launched from position x0 = 0 m and y 0 = 0 m. Its launch velocity, also called the initial velocity, is v0⃗ . Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time t1 with velocity v1⃗ . Its position at this moment is denoted by (x1 , y max ) (x1 , y 1 ) or since it is at its maximum height. The other point, at time t2 with velocity ⃗ v2 , corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x2 , y 2 ) , also known as (xmax , y 2 ) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y 2 = y 0 = 0 m. Part A How do the speeds v0 , v1 , and v2 (at times t0 , t1 , and t2 ) compare? ANSWER: http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 3/21 3/3/2014 Chapter 3 Homework v0 = v1 = v2 >0 v0 = v2 > v1 =0 v0 = v2 > v1 >0 v0 > v1 > v2 >0 v0 > v2 > v1 =0 Correct Here v0 equals v2 by symmetry and both exceed v1 . This is because v0 and v2 include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time t0 . Recall that t0 refers to the instant just after the ball has been launched, so it is still at ground level (x0 = y 0 = 0 m). However, it is already moving with initial velocity v0⃗ , whose magnitude is v0 = 30.0 m/s and direction is θ = 60.0 degrees counterclockwise from the positive x direction. Part B What are the values of the intial velocity vector components vector components a0,x v 0,x and v0,y (both in m/s) as well as the acceleration and a0,y (both in m/s )? Here the subscript 0 means "at time t0 ." 2 Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector v⃗0 points up and to the right. Since "up" is the positive y axis direction and "to the right" is the positive x axis direction, As shown in the figure, v 0,x v0,x , v0,y v0,x and v0,y will both be positive. , and v0 are three sides of a right triangle, one angle of which is θ . Thus and v0,y can be found using the definition of the sine and cosine functions given below. Recall that θ = 60.0 degrees http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 0 = 30.0 m/s 4/21 3/3/2014 Chapter 3 Homework v 0 = 30.0 m/s and θ = 60.0 degrees and note that sin(θ) = cos(θ) = length of opposite side length of hypotenuse length of adjacent side length of hypotenuse = = v0,y v0 v 0,x v0 , . What are the values of v0,x and v0,y ? Enter your answers numerically in meters per second separated by a comma. ANSWER: 15.0,26.0 m/s ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Correct Also notice that at time t2 , just before the ball lands, its velocity components are v2,x as always) and v2,y = −26.0 m/s (the same size but opposite sign from v0,y (the same by symmetry). The acceleration at time t2 will have components (0, -9.80 m/s ), exactly the same as at 2 = 15 m/s t0 , as required by Rule 2. The peak of the trajectory occurs at time t1 . This is the point where the ball reaches its maximum height y max . At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components vector components a1,x v 1,x and v1,y (both in m/s) as well as the acceleration and a1,y (both in m/s )? Here the subscript 1 means that these are all at time t1 . 2 ANSWER: http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 5/21 3/3/2014 Chapter 3 Homework 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (t0 ) until just before it lands (t2 ). Hence the flight time can be calculated as t2 − t0 , or just t2 in this particular situation since t0 = 0 . Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height y max , how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v0,y = 0 ) as well as the same acceleration (a⃗ = g downward). They differ only in their x velocity (one is zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 6/21 3/3/2014 Chapter 3 Homework The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. ANSWER: t0 t1 − t0 t2 t2 − t1 t2 − t0 2 Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 - x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}} {v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}} {t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 7/21 3/3/2014 Chapter 3 Homework Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. Exercise 3.10 A daring 510-{\rm N} swimmer dives off a cliff with a running horizontal leap, as shown in the figure . Part A What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.75 {\rm m} wide and 9.00 {\rm m} below the top of the cliff? ANSWER: v_0 = 1.29 {\rm m/s} http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 8/21 3/3/2014 Chapter 3 Homework Correct Exercise 3.12 A rookie quarterback throws a football with an initial upward velocity component of 15.9{\rm m/s} and a horizontal velocity component of 19.9{\rm m/s} . Ignore air resistance. Part A How much time is required for the football to reach the highest point of the trajectory? Express your answer using three significant figures. ANSWER: t_1 = 1.62 {\rm s} Correct Part B How high is this point? Express your answer using three significant figures. ANSWER: h = 12.9 {\rm m} Correct Part C How much time (after it is thrown) is required for the football to return to its original level? Express your answer using three significant figures. ANSWER: t_2 = 3.24 {\rm s} Correct Part D How does this compare with the time calculated in part (a). http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 9/21 3/3/2014 Chapter 3 Homework Express your answer using three significant figures. ANSWER: \large{\frac{t_2}{t_1}} = 2.00 Correct Part E How far has it traveled horizontally during this time? Express your answer using three significant figures. ANSWER: x = 64.6 {\rm m} Correct Exercise 3.13 A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 22.4{\rm m} above the river, while the opposite side is a mere 1.6{\rm m} above the river. The river itself is a raging torrent 57.0{\rm m} wide. Part A How fast should the car be traveling just as it leaves the cliff in order just to clear the river and land safely on the opposite side? ANSWER: v_0 = 27.7 {\rm m/s} Correct Part B What is the speed of the car just before it lands safely on the other side? ANSWER: v = 34.3 {\rm m/s} Correct http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 10/21 3/3/2014 Chapter 3 Homework Exercise 3.15 Inside a starship at rest on the earth, a ball rolls off the top of a horizontal table and lands a distance D from the foot of the table. This starship now lands on the unexplored Planet X. The commander, Captain Curious, rolls the same ball off the same table with the same initial speed as on earth and finds that it lands a distance 2.43 D from the foot of the table. Part A What is the acceleration due to gravity on Planet X? ANSWER: g_{\rm X} = 1.66 {\rm m/s^2} Correct Exercise 3.19 In a carnival booth, you win a stuffed giraffe if you toss a quarter into a small dish. The dish is on a shelf above the point where the quarter leaves your hand and is a horizontal distance of 2.1 {\rm m} from this point (the figure ). If you toss the coin with a velocity of 6.4 {\rm{ m/s}} at an angle of 60 ^\circ above the horizontal, the coin lands in the dish. You can ignore air resistance. Part A What is the height of the shelf above the point where the quarter leaves your hand? Express your answer using two significant figures. ANSWER: H = 1.5 {\rm m} http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 11/21 3/3/2014 Chapter 3 Homework Correct Part B What is the vertical component of the velocity of the quarter just before it lands in the dish? Express your answer using two significant figures. ANSWER: v_y = -0.89 {\rm m/s} Correct Problem 3.51 A jungle veterinarian with a blow-gun loaded with a tranquilizer dart and a sly 1.5-{\rm kg} monkey are each a height 25{\rm m} above the ground in trees a distance 70{\rm m} apart. Just as the hunter shoots horizontally at the monkey, the monkey drops from the tree in a vain attempt to escape being hit. Part A What must the minimum muzzle velocity of the dart have been for the hunter to hit the monkey before it reached the ground? Express your answer using two significant figures. ANSWER: v = 31 {\rm m/s} Correct Circular Launch A ball is launched up a semicircular chute in such a way that at the top of the chute, just before it goes into free fall, the ball has a centripetal acceleration of magnitude 2 \texttip{g}{g}. http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 12/21 3/3/2014 Chapter 3 Homework Part A How far from the bottom of the chute does the ball land? Your answer for the distance the ball travels from the end of the chute should contain \texttip{R}{R}. Hint 1. Speed of ball upon leaving chute How fast is the ball moving at the top of the chute? Hint 1. Equation of motion The centripetal acceleration for a particle moving in a circle is a_c = v^2/r, where \texttip{v}{v} is its speed and \texttip{r}{r} is its instantaneous radius of rotation. ANSWER: \texttip{v}{v} = \sqrt{\left(2 g R\right)} Incorrect; Try Again; 5 attempts remaining Hint 2. Time of free fall How long is the ball in free fall before it hits the ground? Express the free-fall time in terms of \texttip{R}{R} and \texttip{g}{g}. Hint 1. Equation of motion There is constant acceleration due to gravity, so you can use the general expression \large{y(t) = y_0 + v_{y0}t + \frac{a_y t^2}{2}}. http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 13/21 3/3/2014 Chapter 3 Homework Write the values of \texttip{y_{\rm 0}}{y_0}, \texttip{v_{\rm y0}}{v_y0}, and \texttip{a_{\mit y}}{a_y} (separated by commas) that are appropriate for this situation. Use the standard convention that \texttip{g}{g} is the magnitude of the acceleration due to gravity. Take y = 0 at the ground, and take the positive y direction to be upward. ANSWER: \texttip{y_{\rm 0}}{y_0}, \texttip{v_{\rm y0}}{v_y0}, \texttip{a_{\mit y}}{a_y} = 2 R,0,-g Hint 2. Equation for the height of the ball To find the time in free fall before the ball hits the ground, \texttip{t_{\rm f \hspace{1 pt}}}{t_f}, set the general equation for the height equal to the height of the ground. Answer in terms of \texttip{t_{\rm f \hspace{1 pt}}}{t_f}, \texttip{R}{R}, and \texttip{g}{g}. ANSWER: y(t_{\rm f \hspace{1 pt}})=0 = \large{2 R-\frac{g}{2} {t_{f}}^{2}} ANSWER: \texttip{t_{\rm f \hspace{1 pt}}}{t_f} = \large{2 \sqrt{\frac{R}{g}}} Hint 3. Finding the horizontal distance The horizontal distance follows from D = x(t_{\rm f \hspace{1 pt}}) = x_0 + v_{x0}t_{\rm f \hspace{1 pt}}, where x_0=0. \texttip{v_{\rm x0}}{v_x0} and \texttip{t_{\rm f \hspace{1 pt}}}{t_f} were found in Parts i and ii respectively. ANSWER: \texttip{D}{D} = \large{\sqrt{19.6 R} \sqrt{\frac{4R}{9.8}}} Correct Direction of Acceleration of Pendulum Learning Goal: To understand that the direction of acceleration is in the direction of the change of the velocity, which is unrelated to the direction of the velocity. The pendulum shown makes a full swing from -\pi/4 to + \pi/4. Ignore friction and assume that the string is massless. The eight labeled arrows represent directions to be referred to when answering the following questions. http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 14/21 3/3/2014 Chapter 3 Homework Part A Which of the following is a true statement about the acceleration of the pendulum bob, \texttip{\vec{a}}{a_vec}. ANSWER: \texttip{\vec{a}}{a_vec} is equal to the acceleration due to gravity. \texttip{\vec{a}}{a_vec} is equal to the instantaneous rate of change in velocity. \texttip{\vec{a}}{a_vec} is perpendicular to the bob's trajectory. \texttip{\vec{a}}{a_vec} is tangent to the bob's trajectory. Correct Part B What is the direction of \texttip{\vec{a}}{a_vec} when the pendulum is at position 1? Enter the letter of the arrow parallel to \texttip{\vec{a}}{a_vec}. Hint 1. Velocity at position 1 What is the velocity of the bob when it is exactly at position 1? ANSWER: \texttip{v_{\rm 1}}{v_1} = 0 {\rm m/s} Correct Hint 2. Velocity of bob after it has descended http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 15/21 3/3/2014 Chapter 3 Homework What is the velocity of the bob just after it has descended from position 1? ANSWER: very small and having a direction best approximated by arrow D very small and having a direction best approximated by arrow A very small and having a direction best approximated by arrow H The velocity cannot be determined without more information. Correct ANSWER: H Correct Part C What is the direction of \texttip{\vec{a}}{a_vec} at the moment the pendulum passes position 2? Enter the letter of the arrow that best approximates the direction of \texttip{\vec{a}}{a_vec}. Hint 1. Instantaneous motion At position 2, the instantaneous motion of the pendulum can be approximated as uniform circular motion. What is the direction of acceleration for an object executing uniform circular motion? ANSWER: C Correct We know that for the object to be traveling in a circle, some component of its acceleration must be pointing radially inward. Part D What is the direction of \texttip{\vec{a}}{a_vec} when the pendulum reaches position 3? Give the letter of the arrow that best approximates the direction of \texttip{\vec{a}}{a_vec}. http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 16/21 3/3/2014 Chapter 3 Homework Hint 1. Velocity just before position 3 What is the velocity of the bob just before it reaches position 3? ANSWER: very small and having a direction best approximated by arrow B very small and having a direction best approximated by arrow C very small and having a direction best approximated by arrow H The velocity cannot be determined without more information. Hint 2. Velocity of bob at position 3 What is the velocity of the bob when it reaches position 3? ANSWER: \texttip{v_{\rm 3}}{v_3} = 0 {\rm m/s} ANSWER: F Correct Part E As the pendulum approaches or recedes from which position(s) is the acceleration vector \texttip{\vec{a}}{a_vec} almost parallel to the velocity vector \texttip{\vec{v}}{v_vec}. ANSWER: position 2 only positions 1 and 2 positions 2 and 3 positions 1 and 3 Correct Exercise 3.25 The earth has a radius of 6380 {\rm km} and turns around once on its axis in 24 {\rm h}. http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 17/21 3/3/2014 Chapter 3 Homework Part A What is the radial acceleration of an object at the earth's equator? Give your answer in {\rm{m/s}}^2. ANSWER: a_{\rm rad} = 3.40×10−2 {\rm{m/s}}^2 Correct Part B What is the radial acceleration of an object at the earth's equator? Give your answer as a fraction of {\it g}. ANSWER: a_{\rm rad} = 3.40×10−3 {\it g} Correct Part C If a_{{\rm{rad}}} at the equator is greater than {\it g} , objects would fly off the earth's surface and into space. What would the period of the earth's rotation have to be for this to occur? ANSWER: T = 5070 {\rm s} Correct Exercise 3.29 A Ferris wheel with radius 14.0 {\rm m} is turning about a horizontal axis through its center (the figure ). The linear speed of a passenger on the rim is constant and equal to 6.40{\rm {\rm m/s}} . http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 18/21 3/3/2014 Chapter 3 Homework Part A What is the magnitude of the passenger's acceleration as she passes through the lowest point in her circular motion? ANSWER: a = 2.93 {\rm m/s^2} Correct Part B What is the direction of the passenger's acceleration as she passes through the lowest point in her circular motion? ANSWER: towards the center outwards the center Correct Part C What is the magnitude of the passenger's acceleration as she passes through the highest point in her circular motion? ANSWER: a = 2.93 {\rm m/s^2} http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 19/21 3/3/2014 Chapter 3 Homework Correct Part D What is the direction of the passenger's acceleration as she passes through the highest point in her circular motion? ANSWER: towards the center outwards the center Correct Part E How much time does it take the Ferris wheel to make one revolution? ANSWER: T = 13.7 {\rm s} Correct Exercise 3.30 At its Ames Research Center, NASA uses its large “20-G” centrifuge to test the effects of very large accelerations (“hypergravity”) on test pilots and astronauts. In this device, an arm 8.84 {\rm m} long rotates about one end in a horizontal plane, and the astronaut is strapped in at the other end. Suppose that he is aligned along the arm with his head at the outermost end. The maximum sustained acceleration to which humans are subjected in this machine is typically 12.5 {\it g}. Part A How fast must the astronaut's head be moving to experience this maximum acceleration? ANSWER: v = 32.9 {\rm m/s} Correct Part B http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 20/21 3/3/2014 Chapter 3 Homework What is the difference between the acceleration of his head and feet if the astronaut is 2.00 {\rm m} tall? ANSWER: \Delta a = 27.7 {\rm m/s^2} Correct Part C How fast in rpm \left( {\rm rev/min} \right) is the arm turning to produce the maximum sustained acceleration? ANSWER: \large{\frac{1}{T}} = 35.5 {\rm rpm} Correct Problem 3.88 A projectile is thrown from a point P. It moves in such a way that its distance from P is always increasing. Part A Find the maximum angle above the horizontal with which the projectile could have been thrown. You can ignore air resistance. ANSWER: \phi = 70.5 ^\circ Correct Score Summary: Your score on this assignment is 98.3%. You received 13.77 out of a possible total of 14 points. http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 21/21