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

C

Explanation

Distance travelled by the particle is `x=40+12t-t^(3)` We know that, speed is defined as the rate of change of distance i.e.,
`upsilon=(dx)/(dt)`
`therefore upsilon=(d)/(dt)(40+12t-t^(3))=0+12-3t^(2)`
But final velocity, `upsilon = 0`
`therefore 12-3t^(2)=0rArr t^(2)=(12)/(3)=4 rArr t=2s`
Hence, distance travelled by the particle before coming to rest is given by
`x=40+12(2)-(2)^(3)=40+24-8=64-8=56m`