Solutions Manual Fishbane Physics for Scientists and Engineers 3rd EditionFishbane Instant dowload and all chapters Solutions Manual Fishbane Physics for Scientists and Engineers 3rd Edition Fishbane https://testbankdata.com/download/solutions-manual-fishbane-physics-scientists- engineers-3rd-edition-fishbane/ Chapter 2 Straight-Line Motion Answers to Understanding the Concepts Questions 1. You should be worried about something that might happen to bring the car in front of you to a stop. Your stopping time depends on two factors: Your fixed reaction time, and the time required for your brakes to bring your car to a stop. At a higher initial speed, you travel farther during the time it takes you to react and apply the brakes, and you travel farther in the t ime it takes your brakes to bring you to a stop. Both factors, then, argue for increasing spacing with increasing speed. 2. The velocity of the chalk is zero at that point, but the acceleration remains g, downward. In fact if the acceleration were z ero as well then the chalk would maintain zero velocity there — i.e., it would "freeze" at the top of its path! 3. We have seen that for a fixed acceleration g x , the relation between fall distance d and fall time t is, assuming the falling object starts from rest, d = \g x t 2. Thus for fixed d, t = (2d/gx) 1/2. The variation from planet to planet for t, that is, with g x , is then (gx) - l/2 . The larger g x , the smaller the fall time. The speed of the object at the end of the fall is v = g x t = (2dg x) 1/2. The speed increases with g x like (gx)1/2 . 4. Both you and the bowling balls would be falling at the same rate (g), so there is no reason to worry that any of them would crash onto you. 5. The acceleration of a falling object is equal to g only in a true free fall, which is devoid of any air resistance. In reality, as the object falls, it encounters an air resistance which increases with its speed. Initially, the object is not moving very fast so the air resistance exerted on it is not yet significant, and it falls with an acceleration close to g . As it speeds up, however, it picks up more and more air resistance so its acceleration gradually diminishes, until it reaches a certain speed, at which the upward air resistance equals the downward gravitational pull, whereupon its acceleration is zero and the its speed can no longer increas e. This speed is therefore referred to as the terminal speed. So no, the speed of a falling object cannot increase indefinitely. 6. Certainly if the (negative) acceleration has a constant magnitude, the velocity cannot remain positive. Indeed, if the initia l velocity has magnitude v0, and the acceleration has the constant magnitude a, then the velocity varies with time according to v = v0 - at, and v = 0 at a time t = v0/a; for times greater than this the velocity is negative. However, the acceleration could steadily decrease in magnitude, while remaining negative, such that the velocity could remain positive. A physical example occurs when a rocket is sent away from Earth with enough initial speed to leave the Solar System --we say that its initial speed exceeds the "escape speed." If we say that "up" is the positive direction, then the acceleration is negative while the velocity is positive. As the object moves away, the force of gravity on it, and hence its acceleration, decreases in magnitude while remaining negative. For a fast enough start, the object never comes to rest or turns around. Incidentally, the escape speed from Earth is about 11.2 km/s. 7. Treat the jump of the astronaut as a projectile motion. Then the height he can reach is h = v0 2/2g, which is inversely proportional to g. Since the astronaut can jump 1.2 m on the surface of Earth, assuming that his initial jumping speed does n ot change, then he would be able to jump as much as (0.8 m)[(9.8 m/s 2)/(1.6 m/s2) ] = 5 m on the surface of the Moon. Note that we neglected the height of the Page 2-1 Chapter 2: Straight-Line Motion astronaut. To get a more precise result, we need to find how much the center of mass of the astronaut can rise on the surface of Earth, and multiply that number by [(9.8 m/s 2)/(1.6 m/s 2) ] = 6.1 to obtain the corresponding value on the Moon. 8. Common sense tells us that a sudden decrease in speed, i.e., a large deceleration, can cause damage to our body. The airbag prolongs the deceleration process as the object's speed decreases to zero in a collision. So the magnitude of the deceleration of the object is reduced, lowering the chance of injury or damage. 9. True. This description is consistent with the case when the object is undergoing a uniformly accelerated motion with a pointing to the left, i.e., in the negative x-direction (so a < 0). The velocity of the object as a function of time is v = v 0 + at. Since the object starts out moving to the right v 0 > 0. But since a <0, v will decrease, and at t = - v0 /a we have v = 0, when the object stops. As t further increases v becomes negative, meaning that the object's direction of motion is now to the left, while the magnitude of v increase with time so the object speeds up. 10. As long as the motion is in the positive direction, so that the velocity always is positive, there will be no difference betw een the average speed and the magnitude of the average velocity. This corresponds to Table 2-1. Once negative velocities can occur, even along a straight line, then the average velocity can have any magnitude, including zero, while the average speed is alwa ys greater than zero if there is any movement at all. For more complicated motions, the two quantities are not closely related. For example, when a runner goes exactly once around a track, the average velocity is zero (as the net displacement is zero) while the average speed is not. 11. Suppose that the runner does not have a false start, so he or she cannot start running (from rest) until t = 0. So the initial speed at t = 0 should be zero. This is supported by the distance versus time curve, which shows a slope of zero at t = 0. 12. The ancient Greek mathematicians never learned the admittedly subtle notion of a limit — that the summation of a larger number of smaller and smaller terms, in this case the terms corresponding to the smaller and smaller subdivisions in time and distance traveled, can add to a finite result, in this case the finite time for the runner to catch a tortoise a finite distance ahead of him. Zeno's paradox certainly doesn't correspond to our experience! 13. Suppose that the object in question travels x 0 from to x. Then v is proportional to (x - x0)1/2 , so v2 is proportional to (x - x0). Compare this with the equation v 2 = v02 + 2a (x - x0), and we see that the motion is uniformly accelerated, with zero initial speed (v0 = 0). 14. No. Velocity and acceleration can have different signs. For example, if a car is moving forward, which we choose as the positive direction, then if the driver slams the brake to slow down the car, the acceleration of the car would be negative wh ile its velocity remains positive. In general, if the velocity and acceleration of an object have t he same sign, then it must be speeding up; if they have opposite signs it must be slowing down. 15. False. Suppose that the body falls from rest, starting from the origin of the x-axis which points downward. Then v 2 = 2gx. As the object has fallen twice as far x doubles to 2x, at which time v 2 = 2g(2x) = 4gx. So v2 doubles as x does, and v itself increases only by a factor of /2, not four times. 16. Neglecting the effects of air resistance, all three beanbags have exactly the same constant acceleration, namely the accelera tion of gravity g, all the time they are in the air. There isn't much to compare here. 17. Suppose that the three bags are tossed out at about the same time. Then the third one hits the floor first, followed by the second one, then the first one. The first and the third bag will hit the ground with the same speed, which is higher than tha t of the second one. Page 2-2 Fishbane, Gasiorowicz, and Thornton 18. The beanbag is in free fall, so its acceleration is always g, downward, even though its velocity varies. 19. Let the diameter of each wheel be d. Then the distance covered by the bicycle with each turn of its wheels is nd. If the wheels turn at the rate of N turns/s, then the total linear distance covered by the bicycle per second is N(nd). But by definition this is its linear speed: v = N(nd), which gives N = v / nd. So in addition to v we would also need to measure the diameter of each wheel. 20. Let's assume that the amount of time spent going up equals the amount of time spent coming back down. That means that we can measure the total time of many jumps, an easy thing to do with precision, divide by the total number of jumps to find the tim e of one jump, and then divide by two to find the time At to fall, say. Our meter stick allows us to measure the height h of a jump; with this information, we can find the average acceleration a by applying the formula h = \ a (At) 2, or a = 2h / (At)2 . 21. Here are a few examples: A box sliding up or down a straight ramp, a glider on an air track being pulled by a hanging mass, two unequal masses connected by a string hanging over a fixed pulley, a vehicle accelerating uniformly down a straight road. (Not e: if an object moves in a circle at uniform speed, its acceleration is not a constant — the direction of that acceleration is always changing as the object moves. See Chapter 3.) 22. To measure velocity in a straight-line motion we need to know the distance traveled and the time it takes to travel that distance. With a measuring rod we can determine the distance; and since the speed of the film is given (say, 24 frames a second), we al so know the time interval between any two frames. For example, we can examine two adjacent frames to determine the distance the person travels, and divide this by the time interval between the two frames. This gives us a reasonably good measurement of t he instantaneous velocity of the person, since the time interval between two movie frames is fairly shor t. Once the velocity is measured, we can then find the difference in velocity between, say, a couple of frames, and divide it by the corresponding ti me interval. This gives us the approximate value of the instantaneous acceleration. 23. Choose up as positive. The v vs t graph is shown below, where the slope of the line is always equal to -g. The velocity of the ball is zero again at t = 20 s, when it returns to the point where it was initially dropped. 24. As the ball makes contact with the ground it encounters an upward resistance from the ground, quickly causing its downward acceleration to decrease from g to zero. After that the acceleration of the ball becomes upward and the ball eventually acquires an upward speed before leaving contact with the ground. t t t (a) object at rest (b) object moving slowly (c) object moving rapidly The main disadvantage of switching the axes is that the velocity of the object is no longer equal to the slope of the curve, but rather the reciprocal of the slope. Page 2-3 Chapter 2: Straight-Line Motion Chapter 2: Straight-Line Motion Page 2-7 Chapter 2: Straight-Line Motion 16. (a) x (m) t (s) (b) For the average velocity, we have Ax V(2.0 m/s 2)(t f +1.0 s) ^(2.0m/s 2)(t i +1.0 s) j At tf -1 i v a Between 1.0 s and 5.0 s, this gives AA xx V(2 m/s122) (5.0 s +1.0 s) - V(2.0 m/s 2)(1.0 s +1.0 s) ^(2.0.0 m/s j . v av - AtAt 5.05.0 s -1.0 s s s -1.0 1 i = (0.366 m/s )i . Between 2.0 2.0 ss and and 4.0 4.0 s, s, this this gives gives A x.0 m/s 2)(4.0 s +1.0 s) - V(2.0m/s2)(2.0 s +1.0 s) i i = (0.356 m/s )i . At _ 4.0 s - 2.0 s 1 At 4.0 s - 2.0 s Between 2.8 s and 3.2 s, this gives r A x T/(2 .0 m/s 2) (3.2 s +1.0 s) - i/(2 .0 m/s 2)(28 s +1.0 s) j j i = (0.354 m/s ) . At 3.2 s - 2.8 s (b) For the in stantaneous velocity, V_ 11 lij LUl LIUi IV.V we/ V 2have UO V V.1VV.11J V V_ II dxdx11 dt 2 2.0 m/s i ^( 2 . 0 m/s 2)(t +1.0 s) 1 " “ t +1.0 s) ’ s this is At 3.0 s thisd*is 12.0 m/s 2 j r,n , z „ dx 1 . "'/ ’ ,j — = — ------- 2 =• i = (0 354 m/s )i . dt 2 1.0m/s ^ v(2.0 m/s ) (3.0 )s (3.0 +1.0s s) +. Note that the smaller the At for At centered at t = 3.0 s, the closer the average velocity approaches the instantaneous velocity. 17. Because the acceleration is constant, we have v = v0 + a(t - t 0), or v - v0 = a(t - t 0). To reach 35 mi/h: 35 mi/h = 0 + (0.40 m/s 2)(t - t 0), so 2 3 t - t0 = [(35 mi/h)/(0.40 m/s )](1 h/3600 s)(1.609 x 10 m/mi) = 39 s 18. Because the acceleration is constant, we have v = v0 + a(t - t 0); 5 m/s = 10 m/s + (- 0.3 m/s2)t, which gives t = 17 s 19. Because the acceleration is constant, we have v = v0 + a(t - t 0); (60 mi/h)(1.61 km/mi)/(3600 s/h) = 0+ a(9.0 s), which gives a= 3.0 m/s 2 = 0.30g. Page 2-8 Page 2-9 Chapter 2: Straight-Line Motion Page 2-10
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