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^(-kt)`
D. `(v_(0))/(sqrt(2v_(0)^(2) kt + 1))`

Answer

D

Explanation

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