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_(0)`
C. `v_(0)e^(-k//1)`
D. `(v_(0))/sqrt((2v_(0)^(2)kt+1))`

Answer

D

Explanation

at t=0 velocity `=v_(0)`
now `(dv)/(dt)=-kv^(2)rArr v^(-3)dv=-k dt `
Intergrating both sides
`int_(v_(0))^(v)v^(-3)dv=-kint_(0)^(t)dtrArr [(v^(-3+1))/(-3+1)]_(v_(0))^(v)=k[t]_(0)^(1)`
`-(1)/(2)[(2)/(v^(2))-(1)/(v_(0)^(2))]=-kt`
`(1)/(v^(2))=2kt+(1)/(v_(0)^(2)) rArr (1)/(v^(2))=(2kt v_(0)^(2)+1)/(v_(0)^(2))`
`v=(v_(0))/sqrt(2ktv_(0)^(2)+1)`