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Question

At t=0, a body starts from origin with some initial velocity, The displacement x(m) of the body varies with time t(x) as x=(2/3)t2+16t+2. Find the initial velocity of the body and also find how long does the body take to come to rest? What is the acceleration of the body when it comes to rest?

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Solution

v=dxdt=ddt(23t3+16t+2)=43t+16
At t=0,
v=43×0+16=16 ms1
Putting v=0, we get 43t+16=0t=12s
Hence, the body takes 12 s to come to rest (momentarily).
Now a=dvdt=ddt(43t+16)=43ms2
We see that acceleration is constant, so when the body comes to rest, its acceleration is 4/3 ms2.

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