A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as `alpha=ksqrtv,`where k is a positive constant . At the initial moment the velocity of the point is equal to `V_(0)` . What distance will it take to cover that distance ?

Explanation

Let `t_(0)` be the time in which it comes to a stop.
Given that `-(dv)/(dt)=ksqrtV` `int_(0)^(t_(0))kdt=int_(v_(0))^(0)-(dv)/sqrtv`
`:. Kt_(0)=2sqrtv_(0) :. t_(0)=2/ksqrtv_(0)`
Now to find the distance covered before stopping,
`(dv)/(dt)=(dv)/(ds)(ds)/(dt)=v(dv)/(ds)` But ,`(dv)/(dt)=-ksqrtV,`
`:.v(dv)/(ds)=-ksqrtV ` `:. sqrtvdv=-kds`
`:.int_(v_(0))^(0)sqrtvdv=-int_(0)^(S)k dsrArr s =2/(3k)V_(0)^(3/2)`