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. 1s
B. 2s
C. 4s
D. 8(s)

Answer

B

Explanation

`(dv)/(dt) = -2.5 sqrt(v)`
`int_(6.25)^(0) v^(-1//2)dv = int_(0)^(t) 2.5 dt `
`[(v^(1//2))/(1//2)]_(6.25)^(0)=2.5 [t]_(0)^(t)`
`2[0-sqrt(6.25)]= -2.5 (t-0)`
`2[-2.5]= -2.5t`
`t=2s`