Question

A particle with acceleration a = -kv (where k is constant, and v is velocity) crosses origin with velocity v0. Velocity of the particle at point (l, 0) is

Answer 1

a particle wirh acceleration, a = -kv crosses the origin with velocity $v_0$

we know a relation, $a=v\frac{dv}{dx}$

now, $a=v\frac{dv}{dx}=-kv$

or, $\int\limits^v_{v_0}{v}\,dv=-k\int\limits^l_0{dx}$

or, $\left[\frac{v^2}{2}\right]^v_{v_0}=-k[x]^l_0$

or, $\frac{v^2-v^2_0}{2}=-kl$

or, $v^2-v^2_0=-2kl$

or, $v^2=v^2_0-2kl$

or, $v=\sqrt{{v_0}^2-2kl}$

hence, velocity of the particle at point (l,0) is $v=\sqrt{{v_0}^2-2kl}$