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. `(1)/(2ktV_(0))`
B. `((V_(0))/(1+2kt V_(0)^(2)))`
C. `(V_(0))/(sqrt(1-2k t V_(0)^(2)))`
D. `(V_(0))/(sqrt(1+2kt V_(0)^(2)))`

Answer

D

Explanation

`(dV)/(dt)=-kV^(3)`
`underset(v_(0))overset(v)int V^(-3)dV=-kunderset(0)overset(t)intdt`
`[(V^(-2))/(-2)]_(v_(0))^(v)=-kt`
`[(1)/(V^(2))]_(v_(0))^(v)=2kt`
`(1)/(V^(2))-(1)/(V_(0)^(2))=2kt`
`(1)/(V^(2))-2kt+(1)/(V_(0)^(2))=(2ktV_(0)^(2))/(V_(0)^(2))`
`V=(V)/(sqrt((1-2ktV_(0)^(2)))`