An object moves in a straight line with deceleration whose magnitude varies with deceleration whose magnitude varies with velocity as `3v^(2//3).` If at an initial point, the velocity is 8 m/s, then the distence travelled by the object before it stops is

Options

A. 2 m
B. 4 m
C. 6 m
D. 8 m

Answer

B

Explanation

Given that, a= deceleration `= - 3v^(2//3) " " m//s^(2)`
At initial point (t = 0) velocity = 8 m/s
Distance travelled by the object before it stop is
We know that,
`a=(dv)/(dt)rArr a = (dv)/(ds).(ds)/(dt)`
a = vdv / ds
ads = vdv
`-3v^(2//3)ds = vds`
`rArr ds = - (1)/(3)v^(1//3)dv`
Integrate both sides, we get
`int ds = (-1)/(3) int v^(1//3)dv`
`rArr s = (-1)/(4)v^(4//3) + C ...(i)`
`rArr At " " t = 0,u=8 m//s,s=0`
So, `c=(1)/(4) (8) ^(4//3)=(16)/(4)=4`
Therefore, from Eq. (i)
`s = (-1)/(4)v^(4//3)+4`
Body stops, when v = 0.
So, s = 4 m.