Question

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 :

Answer 1

`(dv)/(dt)=-2.5sqrt(v), (or) (1)/(sqrt(v))dv=-2.5dt`
On integrating within limits
`(v_(1)=6.25 m//s` to `v_(2)=0)`
`int_(v_(1)=6.25 m//s)^(v_(2)=0)v^((-1)/(2))dv=-2.5int_(0)^(t)dt`
`2xx[v^(1//2)]_(6.25)^(0)=-(2.5)t (or)`
`t=(-2xx(6.25)^((1)/(2)))/(-2.5)=2s`