A particle moves along a straight line according to the law `s=16-2t+3t^(3)`, where `s` metres is the distance of the particle from a fixed point at the end of `t` second. The acceleration of the particle at the end of `2s` is

Options

A. `3.6 ms^(-2)`
B. `36ms^(-2)`
C. `36 kms^(-2)`
D. `360 ms^(-2)`

Answer

B

Explanation

Given `s=16-2t+3t^(3)`
`implies(ds)/(dt)=-2+9t^(2)implies(d^(2)s)/(dt^(2))=18t`
Now, the acceletation of the particle at the end of `t=2` is
`((d^(2)s)/(dt^(2)))_(t=2)=18xx2=36 ms^(-2)`