A point moves linearly with deceleration which is given by `dv//dt=-alphasqrt(v)`, where alpha is a positive constant. At the start `v=v_(0)`. The distance traveled by particle before it stops will be

Explanation

`(dv)/(dt)=-alphav^(1//2)`
`int_(v_(0))^(v)v^(-1//2) dv=-alphaint_(0)^(t) dt`
`(|v^(-1/2+1)|_(v_(0))^(v))/((-1/2+1))=-alpha|t|_(0)^(t)`
`|v^(1/2)|_(v_(0))^(v)=alpha/2t`
`sqrt(v)-sqrt(v_(0))=-alpha/2timpliesv=[sqrt(v_(0))-(alphat)/2]^(2)`
The particle will stop when `v=0`
`t=(2sqrt(v_(0)))/alpha`
To calculate the displacement by velocity is lengthy.
`(dv)/(dt)=v(dv)/(ds)=-alpha v^(1//2)`
`int_(v_(0))^(v) v^(1//2) dv=-alphaint_(0)^(s) ds`
`| v^(3//2)|_(v_(0))^(v)/(3/2)=-alphas`
`v^(3//2)-v_(0)^(3//2)=-3/2alphas`
When particle stops, `v=0`
`impliess=(2v_(0)^(3//2))/(3alpha)`