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^(v_0) kdt =int_(v_0)^0 -(dv)/sqrtv`
`because kt_0=2sqrtv_0`
`therefore t_0=2/k sqrtv_0`
Now to find the distance covered before stopping ,
`(dv)/(dt)=(dv)/(ds) (ds)/(dt)=v (dv)/(ds)` But , `(dv)/(dt)=-k sqrtV`
`therefore v (dv)/(ds)=-ksqrtV " " therefore sqrtvdv=-kds`
`therefore int_(v_0)^0 sqrtvdv=-int_0^s k ds rArr s = 2/(3K) V_0^(2/3)`