The deceleration exerienced by a moving motor blat, after its engine is cut-off is given by `dv//dt=-kv^(3)`, where `k` is constant. If `v_(0)` is the magnitude of the velocity at cut-off, the magnitude of the velocity at a time `t` after the cut-off is.

Options

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

Answer

D

Explanation

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