The motion of a particle is described as `S =2-3t+4t^(3)` what is the acceleration of the paricle at the point where its velocity is zero ?

Options

A. 0
B. 4 unit
C. 8 unit
D. 12 unit

Answer

D

Explanation

Given rule is :
Distance `S=2 -3t +4t^(3)`
`rarr velocity =(ds)/(dt)=-3+12t^(2)`
`rarr` Acceleration `=(d^(2)s)/(dt^(2))=24t` ltrbgt Since velocity is zero
`therefore (ds)/(dt)=0`
Acceleration (when velocity is zero)
`rarr (d^(2)s)/(dt^(2))=24t =24xx1/2=12` unit