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. (a) v_0/(sqrt ( 1- 2 k v^2 _0t)`
B. (b() v_0/(sqrt(1 + 2 k v_0 t)`
C. (c ) ` v_0 /( sqrt 1 - 2 k v^2 t)`
D. (d) v_0/(sqrt (1 + 2 k v_0 t)`

Answer

B

Explanation

Here, ` (dv)/(dt) =- k v^3 ` or `(dv)/v^3 =- k dt`
Integrating wighin the conditions of motion, we have
` int _v_0^v v^3 (dv) =- int_0 ^t k dt ` or [ 1/ ( 2 v^2]_v_0^v =- k [t]_0^t`
` 1/v^2 - 1/v_0^2 =2 kt ` or ` v = v_0 /(sqrt (1 +2 k v_0^2t)` .