The retardation experienced by a movign motr bat after its engine is cut-off , at the instant (t) is given by`, ` =-k v^4` , where (k) is a constant. If ` v_0` is the magnitude of velocity at the cut-off , the magnitude fo velocity at time (t) after the cut-off is .

Options

A. (a) v_0 /(sqrt( 3 kt v_0^3)`1
B. (b) v_0/(sqrt (3 kt v_0^3 + 1 ^(1//3))`
C. (c ) ` sqrt ( 3 kt v_0^3)
D. (d) (3 kt v_0^3 + 1 )^(1//3)`

Answer

B

Explanation

` a= (dv)/(dt) =- k v^4 ` or ` - ( dv)/ d v^4) = dt`
Integrating both the sides, we have
` - 1/2 int_v_0^v v^(-4) dv - int _0^t dt`
or ` -1/k [ v^(-3)/(-3) ]__0^v = t `or 1/(3k)[1/v^3 - 1/v_0^3 ]=t`
or ` - 1/v^3 = 3 kt + 1/ v_0^3 = ( 3 kt v_0^3 + 1)/v_0^3`
or ` v= v_0/((3 kt v_0^3 +1)^(1//3)`.