A particle moves along a straight line `OX`. At a time `t` (in seconds) the distance `x` (in metre) of the particle is given by `x = 40 +12 t - t^3`. How long would the particle travel before coming to rest ?

Options

A. 24m
B. 40m
C. 56m
D. 16m

Answer

D

Explanation

`x=40+12t-t^(3)`
`v=(dx)/(dt)=0+12-3t^(2)`
for `v=0 rArr 12-3t^(2)=0`
`rArr t^(2)=4 rArrt=2` sec
`rArr x_(t=0)=40` m
`x_(t=2S)=40+12xx2-(2)^(2)=56`
dist. T. `=x_(t=2S)-X_(t=0)-X_(t=0)=56-40=16m`