A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as alpha=ksqrt(v), 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?

A point moves rectilinearly with deceleration whose modulus depends on

1.	A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as alpha=ksqrt(v), 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?
Answer» Solution :Let `t_(0)` be the TIME in which it COMES to a stop. Given that `-(dv)/(dt)=ksqrt(V)` `underset(0)overset(t_(0))intkdt=underset(v_(0))overset(0)int-(dv)/(sqrt(v))` `because kt_(0)=2sqrt(v_(0))` `therefore t_(0)=(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(v)dv=-kds` `therefore underset(V_(0))overset(0)intsqrt(v)dv=-underset(0)overset(s)intkdsimpliess=(2)/(3k)V_(0)^((3)/(2))`