A winding drum raises a cage through a height of 63.0 m. The cage has, at first, an acceleration of 12.0 m/s2 until the velocity of 15.0 m/s is reached, after which the velocity is constant until the cage nears the top, when the final retardation is 5.0 m/s2. Find the time taken for the cage to reach the top in second.
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1 answer

stage 1:
a = 12
v = 0 + 12 t
h = 0 + 0 t + 6 t^2
so
15 = 0 + 12 t
t = 15/12 = 5/4
h = 0 + 0 * 5/4 + 6 (25/16) = 9.375 meters

Stage 2:
a = 0
v = 15 + 0 t = 15
h = 9.375 + 15 t + 0 = 9.375 + 15 t where t is time in stage 2

stage 3
a =  5.0 m/s^2
v = 15  5 tt where tt isttime in stage 3
0 = 15  5 tt
so tt = 3 seconds to stop
63= (9.375 + 15 t) + 15 tt  (1/2)5 tt^2
63 = 9.375 + 15 t + 45  (5/2)9
63 = 9.375 + 15 t + 22.5
t = 2.075 seconds at 15
so add 5/4 + 2.075 + 3 for total time 👍
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answered by Anonymous
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