CHAPTER 3 SOLUTIONS TO PROBLEMS AND QUESTIONS

Citation: Jacqueline D. Spears and Dean Zollman, Instructor's Guide for The Fascination of Physics, (The Benjamin /Cummings Publishing Company, Inc., Menlo Park, CA 1985). Permission granted by the publisher.

A. Review of Chapter Material

A1. A different description will occur when the two people are in reference frames which are moving relative to one another.

A2. Phrases such as "the sun sets" ignore the motion of the earth. Highway speeds are always quoted as if the earth were not moving.

A3. In describing motion, any reference frame which is not accelerating will provide an equally valid description.

A4. The relative velocity between two objects remains unchanged as we change among non-accelerating reference systems.

A5. Subtract the velocity of X from the velocity of Y. In doing so, one must be certain that the directions of the velocities are included.

A6. The principle of relativity is: The laws of physics in any reference frame are the same as those in any other reference frame which is moving at a constant velocity relative to the first one.

B. Using the Chapter Material

B1. (a) The book fell straight down to the floor of the bus. (b) The reference frame was the bus. It was moving relative to the earth. (c) The motion of the book relative to the earth was ignored. (d) The book traveled in the direction that the bus was moving and simultaneously fell toward the floor of the bus. (e) The descriptions differ because the reference frames were moving relative to each other.

B2. (a) If the camera had been attached to the pole vaulter, the photograph would have contained multiple images of the cross bar. In this reference frame the cross bar would be moving closer to the camera. (b) The reference frame attached to the earth (Figure 2-9) would be more useful because we actually see his motion. In the reference frame attached to the pole vaulter we have to infer his motion from the motion of the cross bar.

B3. In a reference frame attached to the tree, the bicycle is moving toward the tree, and the tree is stationary. (b) In a reference frame attached to the bicycle, the bicycle is stationary and the tree is moving toward the bicycle.

B4. (a) To calculate the magnitude of the velocity consider the location of the images of the bicycle's rear wheel. It moves from O meters to 60 meters in eight seconds. So, the magnitude of the velocity = 60 m/8 s = 7.5 m/s. The velocity is (7.5 m/s, forward), where forward is the direction toward the front of the bicycle. (b) The velocity of the tree relative to the bicyclist will be (7.5 m/s, backward). (c) The magnitude will be the same because in each case the reference system is connected to one of the two objects involved. The magnitude could be different if a reference frame which was moving relative to both of the tree and the bicyclist was being used.

B5. Relative velocity = velocity of the airplane relative to earth - velocity of the wind relative to the earth = (400 km/h, east) - (50 km/h, east) = 350 km/h, east

B6. Relative velocity = velocity of police car relative to the earth - velocity of other car relative to the earth = (100 km/hr, direction A) - (50 W hr, direction B). In this case we do not know the directions except that they are opposite to each other. So, direction A = -direction B. Then, relative velocity = (100 km/hr + 50 km/hr, direction A) = (150 km/hr, direction A).

B7. Velocity of David relative to Maria = velocity of Maria relative to the earth - velocity of David relative to the earth = (4 m/s, east) - (3 m/s, east) = 1 m/s, east. David cannot catch up with Maria because she is going faster than he is. She moves 1 meter further in front of him each second that they run.

B8. (a) Velocity of Kim relative to Kevin = Velocity of Kim relative to earth - Velocity of Kevin relative to earth = (6 m/s, east) - (6m/s, west) = (6 m/s, east) + (6 m/s, east) = 12 m/s, east. (b) Kim and Kevin will bump into each other in all reference systems. The principle of relativity states that the fundamental physics is the same in all reference frames moving at a constant velocity relative to one another. If Kim and Kevin bump into each other in one reference frame, they will do so in all other reference frames moving at constant velocities relative to one another.

B9. To determine if he can overtake the speeding car, the police officer must use the relative velocity between himself and the other car. The magnitude of this velocity i8 the speed at which he will catch up with the other car. If this relative speed is very small, it will take a very long time for him to catch up. If the speed is negative he will be "losing ground" to the speeder; and he will never catch up. Then, the police officer will certainly need to radio ahead.

C. Extensions to New Situations

C1. (a) The church tower is not moving. (b) The stars are moving in a circle about a point in the center of the photograph. (c) The reference frame is that of the camera which is located in front of the tower. We have assumed that this reference frame is not moving. (d) This person used the stars as a fixed reference frame. (e) The difference between the two reference frames is the objects which are chosen to be stationary. (f) Both reference frames are equally valid for describing this or any other motion.

C2. (a) In coordinate system A the ball moved a distance of 10 m - 2 m = 8 m. The speed = 8 m/4 s = 2 m/s. (b) In system B the distance moved is 18 m - 10 m = 8 m. Speed = 8 m/4 s = 2 m/s. (c) The speeds are the same. This implies that the two coordinate systems are not moving relative to one another. If they had been moving relative to one another, the ball's velocity would have been different in the two reference frames.

C3. (a) The reference frame in which this photograph was taken was probably not moving relative to the earth. We cannot be certain about this statement because we see no other objects in the picture. However, the ball's motion looks similar to that of balls we observe when seeing them in a system fixed relative to the earth. (b) If the camera had been moving horizontally at the same speed as the ball, the ball's horizontal speed would have been zero relative to the camera. Thus, we would see only the ball's vertical motion. The path followed by the ball would look like this:See Figure

(c) In a reference frame fixed relative to the earth we would see this same motion when we threw the ball straight up. (d) When a ball is dropped in a moving reference frame, its motion must be considered relative to the motion of the reference frame. On earth, both the ball and the building are moving eastward with a rotation of the earth. Relative to one another, they are not moving horizontally. When we drop the ball, it moves only vertically relative to the building. Consequently, it lands beneath the spot from which it was dropped.

C4. (a) The velocity measured by the radar is equal to the velocity of your car relative to the police car = velocity of your car relative to the road - velocity of police car relative to the road = (80 km/h, north) - (70 km/h, south) = 150 km/h, north. The speed measured by the radar is 170 km/h. (b) In this case we know the speed of the cars relative to each other but not the speed Or one Or the cars relative to the road. The equation is: (150 km/hr, direction A) = (50 km/h, direction A) - (x, direction B). Since direction A = - direction B, the equation for the magnitudes becomes: 150 km/h = 50 km/h + x which gives x = 100 km/h. (c) In this case the cars are moving in the same direction 80 the equation for the magnitudes of the velocities is: 0 km/h = 90 km/h- x. So, the speed, x, = 90 km/h.

C5. When the airplane is traveling from New York to San Francisco, it is traveling in the opposite direction that the earth is rotating. The trip from San Francisco to New York is in the same direction as the earth is moving. Thus, the speed of the airplane relative to the earth is greater on the New York-San Francisco trip than it is on the San Francisco-New York trip. A trip from New York to San Francisco would be faster if this effect were the only consideration.

C6. (a) Velocity of the bicyclist relative to the wind = velocity of bicyclist relative to the earth - velocity of the wind relative to the earth = (10 km/h, east) - (10 km/h, east) = O. (b) Relative velocity = (10 km/h, east) - (10 km/h, west) = 20 km/h, east. (c) The cyclist will feel the wind moving past her only in the situation described in part b. For the situation in part a, she and the wind are not moving relative to one another, so she will feel no wind. (d) When riding with the wind, the cyclist will not have the wind moving past her and evaporating sweat. If she and the wind are not moving relative to one another, then the situation is similar to riding an exercise bike in a closed room. The cooling effect of the wind will not be present.

C7. (a) In one second the bus moves (30 m/s) x 1 s = 30 m relative to the earth. (b) You move (2 m/s) x 1 s = 2 m. (c) You have moved 30 meters forward with the bus and at the same time 2 meters backward inside it. Relative to the earth you have moved 30 meters - 2 meters = 28 meters, forward. (d) Your velocity relative to the earth will be (net displacement)/time = 28 meters, forward/1 s = 28 m/s, forward. (e) Based on this example we can conclude that we subtract the magnitudes when in opposite directions. This implies that we add the velocities as vectors. To see if this works for velocities in the same direction try problem C8.

C8. (a) Velocity relative to the earth = velocity relative to still water + velocity of water relative to the earth = (2 m/s, direction A) + (1 m/s, direction A) - 3 m/s, direction A. (b) Velocity relative to the earth = (2 m/s, direction A) + (1 m/s, direction A) - 1 m/s, direction A. This result fits with our common sense. When you paddle with the current, your speed relative to the earth will be greater than when you paddle against the current. Common sense and the equation in problem C8 give the same result.

C9. [NOTE: A misprint appeared in this problem for the first printing of the book. The velocity of the antelope should have been 11 m/s, west. This error was corrected in subsequent printings so most students should have the corrected version of the problem.] (a) Relative velocity = (23 m/s, west) - (11 m/s, west) = 12 m/s, west. The lion will be able to catch the antelope. (b) This reference frame must be moving relative to the earth. In fact, because the magnitudes of the velocities are higher than those relative to the earth, the reference frame must be moving east. Its speed is (100 km/h - 23 km/h) = 77 km/h. (c) The relative velocity will be the same for both reference frames even though the reference frames are moving relative to each other. Thus, from a comparison the relative velocity in each frame only, we could not determine if one were moving relative to the other.

C10. To complete this problem we rearrange the relative velocity equation to read: Relative velocity = Relative Velocity + Relative velocity of A to earth of A to B of B to Earth

(a) velocity of car A relative to earth = 30 km/h, direction X + 70 km/h, direction X = 100 km/h, direction X. The actual speed relative to the road is 100 km/h. (b) The measured speed relative to the road is 30 km/h + 60 km/h = 90 km/h. (c) Car B is moving in a direction opposite to the police car, so we subtract the relative speeds. Speed of car B relative to the road = 160 km/h - 70 km/h = 90 km/h. (d) The measured speed of car B relative to the road = 160 km/h - 60 km/h = 100 km/h. (e) Car A is actually speeding. However, because of the cosine error the driver of car B will receive a speeding ticket while the driver of car A will not. (f) To use the cosine error as a court defense you need to be traveling in a direction opposite to that of the police car. Since this error requires an incorrect speed of the police car, it will not serve as a defense if the police were not moving.