# 5 m s 75 m s 15 m s x 50 s 30 s 32

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Unformatted text preview: his time is t = = = 20 s . Thus, the train is slowing down for the a −1.0 m s 2 first 20 s and is at rest for the last 20 s of the 40 s interval. The acceleration is not constant during the full 40 s. It is, however, constant during the 1 first 20 s as the train slows to rest. Application of ∆x = v i t + at 2 to this interval gives the 2 stopping distance as ∆x = ( 20 m s ) ( 20 s ) + 2.36 ( ) 1 2 −1.0 m s 2 ( 20 s ) = 200 m . 2 v f = 40.0 m i h = 17.9 m s an d v i = 0 (a) To find the distance traveled, we use ⎛ v f + vi ⎞ ⎛ 17.9 m s + 0 ⎞ ∆x = v t = ⎜ ⎟ (12 .0 s ) = 107 m . ⎟t =⎜ ⎝ ⎠ 2 ⎝2⎠ (b) The acceleration is a = v f − vi t = 17.9 m s − 0 = 1.49 m s 2 . 12.0 s 32 CHAPTER 2.37 2 At the end of the acceleration period, the velocity is ( ) v f = v i + at = 0 + 1.5 m s 2 ( 5.0 s ) = 7.5 m s . This is also the initial velocity for the braking period. ( ) (a) After braking, v f = v i + at = 7.5 m s + − 2 .0 m s 2 ( 3.0 s ) = 1.5 m s . (b) The total distance traveled is ∆x = ( ∆x ) accel + ( ∆x )brake = ( v t ) accel + ( vt )brake ⎛ 0 + 7.5 m s ⎞ ⎛ 7.5 m s + 1.5 m s ⎞ ∆x = ⎜ ⎟ ( 5.0 s ) + ⎜ ⎟ ( 3.0 s ) = 32 m . ⎝ ⎠ ⎝ ⎠ 2 2 2.38 The initial velocity of the train is v i = 82 .4 k m h and the final velocity is v f = 16.4 km h . The time required for the 400 m train to pass the crossing is found from ⎛ v f + vi ⎞ t as ∆x = vt = ⎜ ⎝2⎟ ⎠ t= 2.39 2 ( ∆x ) 2 ( 0.400 k m ) ⎛ 3600 s ⎞ 29.1 s = ⎜ ⎟= v f + v i ( 82 .4 + 16.4) k m h ⎝ 1 h ⎠ (a) Take t = 0 at the time when the player starts to chase his opponent. At this time, the opponent is 36 m in front of the player. At time t > 0 , the displacements of the players from their initial positions are ( ) 1 1 x player = ( v i ) player t + aplayer t 2 = 0 + 4.0 m s 2 t 2 , 2 2 (1) 1 and, ∆x opponent = ( v i ) opponent t + aopponent t 2 = (12 m s ) t + 0 2 (2) When the players are side-by-side, ∆x player = ∆x opponent + 36 m (3) From Equations (1), (2), and (3), we find t 2 − ( 6.0 s ) t − 18 s 2 = 0 which has solutions of t = − 2.2 s and t = + 8.2 s . Since the time must be...
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