CHAPTER 2 - 20

# CHAPTER 2 - 20 - The negative sign indicates the downward...

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0 0.5 1.0 1.5 2.0 –0.5 –1.0 –1.5 –2.0 10 20 30 40 50 t  (s) v  (m/s) (6.0 m/s) t   = 8.0 m + (8.0 m/s)( t  – 2.0 s), which gives  t  =       4.0 s . ( b ) We can find the distance from the motion of the train: x train  =  v train t  = (6.0 m/s)(4.0 s) =       24 m . 83. We use a coordinate system with the origin at the top of the cliff and up positive. ( a ) For the motion of the stone from the top of the cliff to the ground, we have y  =  y 0  +  v 0 t  +  ! at 2 ; – 65.0 m   = 0 + (10.0 m/s) t   +  ! (– 9.80 m/s 2 ) t 2 . This is a quadratic equation for  t , which has the solutions  t  = – 2.76 s, 4.80 s. Because the stone starts at  t  = 0, the time is        4.80 s . ( b ) We find the speed from v   v 0  +  at   = 10.0 m/s + (– 9.80 m/s 2 )(4.80 s) = – 37.0 m/s.

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Unformatted text preview: The negative sign indicates the downward direction, so the speed is 37.0 m/s . ( c ) The total distance includes the distance up to the maximum height, down to the top of the cliff, and down to the bottom. We find the maximum height from v 2 = v 2 + 2 ah ; 0 = (10.0 m/s) 2 + 2(– 9.80 m/s 2 ) h , which gives h = 5.10 m. The total distance traveled is d = 5.10 m + 5.10 m + 65.0 m = 75.2 m . 84. The instantaneous velocity is the slope of the x vs. t graph: Page * 10 20 30 40 50 t (s) 10 20 x (m) 85. We use a coordinate system with the origin where the Page *...
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