The deceleration experienced by a moving motor boat after its engine is cut off is given by `dv/dt` =`kv^3`, where is a constant. Ifv, is the magnitude of the velocity at cut off, the magnitude of velocity at time after the cut off is :

Options

A. `(v_0)/(2)`
B. `v_0`
C. `(v_0)/(sqrt(2v_(0)^(2)kt+1))`
D. `v_(0)e^(-kt)`

Answer

c

Explanation

Here `(dv)/(dt)=-kv^(3)` or `int_(v_0)^(v) v^(-3)dv= -kint_(0)^(t)dt`
`[1/2v^2]_(v_0)^(v)=kt` or `(1)/(2v^2)-(1)/(2v_(0)^(2))=kt`
or `v_(0)^(2)-v^(2)=2kt v^(2)v_(0)^(2)`
`v^(2)=(v_(0)^(2))/(2ktv_(0)^(2)+1)` or `v=(v_0)/(sqrt(2ktv_(0)^(2)+1)`