56_PartUniversity Physics Solution

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

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2-26 Chapter 2 EXECUTE: (a) For 0 t = to 5.0 s, 0 0 0 30.0 m/s (5.0 m/s) 75.0 m 22 xx vv 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 −+ = . 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 = . 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(2 0 . 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 vt is not really vertical at 5.00 s. EVALUATE: 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. IDENTIFY and SET UP: 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. EXECUTE: For the first 5.0 s of the motion, 0 0, x v = 5.0 s. t = 0 x vv a t =+ gives (5.0 s). va = This is the initial speed for the second 5.0 s of the motion. For the second 5.0 s: 0 (5.0 s), = 5.0 s, t = 0 150 m. 2 1 00 2 x x x xv t a t −= + gives 150 m (25 s ) (12.5 s ) x x aa and 2 4.0 m/s x a = Use this x a and consider the first 5.0 s of the motion: 2 11 0 (4.0 m/s )(5.0 s) 50.0 m. xx v t = EVALUATE: 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. IDENTIFY: Apply 2 1 2 x x x 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. SET UP: 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 =− . EXECUTE: (a) 2 1 2 x x x t for each object gives 2 1 PP P 2 x at and FF 200 m x . Setting PF x x = gives 2 1 F 2 200 m + . (0.0500 m/s ) (10.0 m/s) 200 m 0 tt . 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) 22 1 P 2 (25.0 m/s)(22.5 s) ( 0.100 m/s )(22.5 s) 537 m x =+ = . The passenger train moves 537 m before the collision. The freight train moves (15.0 m/s)(22.5 s)
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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

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