An object , moving with a speed of ` 6.25 m//s `, is decelerated at a rate given by :
`(dv)/(dt) = - 2.5 sqrt (v)` where `v` is the instantaneous speed . The time taken by the object , to come to rest , would be :

Options

A. `1 s`
B. ` 2 s `
C. ` 4s`
D. ` 8 s`

Answer

B

Explanation

we are given ` (dv)/(dt) =-2.5 sqrt v or (1)/(sqrt( v) dv =-2.5 dt `
on integrating , within limit (at t=0 ` ,v_1 = 6.25 ms^(-1) and ` at any time `v_2 =0) ` , we get
` int _( 6.25 ms^(-1) ) ^(0) v^(-1//2 ) dv =-2.5 int _0^(t) dt , 2xx [v^(1//2) ]_(6.25)^(0) =-( 2.5 ) rArr t = ( -2xx( 6.25)^(1//2))/( -2.5) =2s `