A particle is moving along x-direction with a constant acceleration a. The particle starts from `x=x_0` position with initial velocity u. We can define the position of the particle with time by the relation
`x=x_0+ut+(1)/(2)at^2`
plot the position of the particle in relation with time is following situations
(i) If initial position of the particle is on negativ x-axis, initial velocity is positive and acceleration is negative.
(ii) If initial position is positive, initial velocity is negative and acceleration is positive.

Explanation

Let `t_(0)` be the time which it come to a stop.
Given that `-(dt)/(dt)=ksqrt(v)" "underset(0)overset(t_(0))int kdt=underset(v_(0))overset(0)int-(dv)/(sqrt(V))`
`because kt_(0)=2sqrt(v_(0)) therefore=(2)/(k)sqrt(v_(0))`
Now to find the distance covered before stopping,
`(dv)/(dt)=(dv)/(ds)(ds)/(dt)=v(dv)/(ds)` But, `(dv)/(dt)=-Ksqrt(V),`
`therefore v(dv)/(ds)=-ksqrt(V) therefore sqrt(dv)=-kds`
`therefore underset(v_(0))overset(0)intsqrt(v_(0))int sqrt(v)dv=-underset(0)overset(s)int` k ds`implies s(2)/(3k)V_(0)^((3)/(2))`