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)/sqrt(1+2ktv_(0)^(2))`
B. `v_(0)/(2k)`
C. `v_(0)/sqrt(1+2kt)`
D. `v_(0)/sqrt(2kt)`

Answer

A

Explanation

`(dv)/(dt)=-kv^(3)`
`int_(v_(0))^(v)v^(-3)dv=-kint_(0)^(t) dt`
`|v^(-2)/-2|_(v_(0))^(v)=-k|t|_(0)^(t)`
`|1/v^(2)|_(v_(0))^(v)=2k(t-0)`
`1/v^(2)-1/v_(0)^(2)=2kt`
`1/v^(2)=1/v_(0)^(2)+2kt=(1+2kt v_(0)^(2))/v_(0)^(2)`
`v=v_(0)/sqrt(1+2kt v_(0)^(2))`