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A particle moves along a straight line. Its position at any instant is given by `x = 32t-(8t^3)/3` where x is in metres and t in seconds. Find the acceleration of the particle at the instant when particle is at rest.

A

`-16 ms^-2`

B

`-32 ms^-2`

C

`32 ms^-2`

D

`16 ms^-2`

Text Solution

Verified by Experts

The correct Answer is:
B
`v=(dx)/(dt)=32-8t^2`
`v=0` at `t=2s`
`a=(dv)/(dt)=-16t`
at `t=2s, a=032m//s^2`
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