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. `3.6m//s^(2)`
B. `36m//s^(2)`
C. `36km//s^(2)`
D. `360 m//s^(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 acceleratio of the particle at the end of `t=2` s is
`f=(d^(2)s)/(dt^(2))=18xx2=36m//s^(2)`