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56_PartUniversity Physics Solution

56_PartUniversity Physics Solution - 2-26 Chapter 2 EXECUTE...

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2-26 Chapter 2 E XECUTE : (a) For 0 t = to 5.0 s, 0 0 0 30.0 m/s (5.0 m/s) 75.0 m 2 2 x x v v x x t + + = = = . The ball travels a distance of 75.0 m. For 5.0 s t = to 20.0 s, 0 20.0 m/s 0 (15.0 m/s) 150.0 m 2 x x + = = − . The total distance traveled is 75.0 m 150.0 m 225.0 m + = . (b) The total displacement is 0 75.0 m +( 150.0 m) 75.0 m x x = = − . The ball ends up 75.0 m in the negative x - direction from where it started. (c) For 0 t = to 5.0 s, 2 30.0 m/s 0 6.00 m/s 5.0 s x a = = . For 5.0 s t = to 20.0 s, 2 0 ( 20.0 m/s) 1.33 m/s 15.0 s x a − − = = + . The graph of x a versus t is given in Figure 2.64. (d) The ball is in contact with the floor for a small but nonzero period of time and the direction of the velocity doesn’t change instantaneously. So, no, the actual graph of ( ) x v t is not really vertical at 5.00 s. E VALUATE : For 0 t = to 5.0 s, both x v and x a are positive and the speed increases. For 5.0 s t = to 20.0 s, x v is negative and x a is positive and the speed decreases. Since the direction of motion is not the same throughout, the displacement is not equal to the distance traveled. Figure 2.64 2.65. I DENTIFY and S ET U P : Apply constant acceleration equations. Find the velocity at the start of the second 5.0 s; this is the velocity at the end of the first 5.0 s. Then find 0 x x for the first 5.0 s. E XECUTE : For the first 5.0 s of the motion, 0 0, x v = 5.0 s. t = 0 x x x v v a t = + gives (5.0 s). x x v a = This is the initial speed for the second 5.0 s of the motion. For the second 5.0 s: 0 (5.0 s), x x v a = 5.0 s, t = 0 150 m. x x = 2 1 0 0 2 x x x x v t a t = + gives 2 2 150 m (25 s ) (12.5 s ) x x a a = + and 2 4.0 m/s x a = Use this x a and consider the first 5.0 s of the motion: 2 2 2 1 1 0 0 2 2 0 (4.0 m/s )(5.0 s) 50.0 m. x x x x v t a t = + = + = E VALUATE : The ball is speeding up so it travels farther in the second 5.0 s interval than in the first. In fact, 0 x x is proportional to t 2 since it starts from rest. If it goes 50.0 m in 5.0 s, in twice the time (10.0 s) it should go four times as far. In 10.0 s we calculated it went 50 m 150 m 200 m, + = which is four times 50 m. 2.66. I DENTIFY : Apply 2 1 0 0 2 x x x x v t a t = + to the motion of each train. A collision means the front of the passenger train is at the same location as the caboose of the freight train at some common time. S ET U P : Let P be the passenger train and F be the freight train. For the front of the passenger train 0 0 x = and for the caboose of the freight train 0 200 m x = . For the freight train F 15.0 m/s v = and F 0 a = . For the passenger train P 25.0 m/s v = and 2 P 0.100 m/s a = − . E XECUTE : (a) 2 1 0 0 2 x x x x v t a t = + for each object gives 2 1 P P P 2 x v t a t = + and F F 200 m x v t = + . Setting P F x x = gives 2 1 P P F 2 200 m v t a t v t + = + . 2 2 (0.0500 m/s ) (10.0 m/s) 200 m 0 t t + = . The quadratic formula gives ( ) 2 1 10.0 (10.0) 4(0.0500)(200) s (100 77.5) s 0.100 t = + ± = ± . The collision occurs at 100 s 77.5 s 22.5 s t = = . The equations that specify a collision have a physical solution (real, positive t ), so a collision does occur.
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Motion Along a Straight Line 2-27 (b) 2
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