A ball starts from rest and accelerates at 0.475 m/s 2 while moving down anincli

Question

A ball starts from rest and accelerates at 0.475 m/s2 while moving down aninclined plane 8.45 m long. When itreaches the bottom, the ball rolls up another plane, where, aftermoving 15.45 m, it comes to rest. (a) What is the speed of the ball at the bottomof the first plane?
m/s
(b) How long does it take to roll down the first plane?
s
(c) What is the acceleration along the second plane?
m/s2
(d) What is the ball's speed 7.80 malong the second plane?
m/s (a) What is the speed of the ball at the bottomof the first plane?
m/s
(b) How long does it take to roll down the first plane?
s
(c) What is the acceleration along the second plane?
m/s2
(d) What is the ball's speed 7.80 malong the second plane?
m/s

Explanation / Answer

a) vi = 0, vf2 - 0 = 2as =2*0.475m/s2 *8.45m vf = [2*0.475m/s2 *8.45m ] = 2.83m/s b) d = 8.45m = (1/2)at2 t = (2*8.45m / 0.475 m/s2)=5.96 s c) let it be a we have: vi = 2.83 m/s , vf = 0, s= 15.45m vf2-vi2 =0-(2.83m/s)2 = 2as a = -(2.83m/s)2 /2s = -0.26 m/s2 t = (2*8.45m / 0.475 m/s2)=5.96 s c) let it be a we have: vi = 2.83 m/s , vf = 0, s= 15.45m vf2-vi2 =0-(2.83m/s)2 = 2as a = -(2.83m/s)2 /2s = -0.26 m/s2

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