Explanation :
It is given that,
initial speed of particle is [tex]v_1[/tex]
final speed of particle is [tex]v_2[/tex]
acceleration of particle is [tex]a=cv[/tex]
Let s is the distance traveled by particle before it reaches [tex]v_2[/tex]
Now, using equation of motion as :
[tex]v^2_2-v^2_1=2as[/tex]
so, [tex]s=\frac{v^2_2-v^2_1}{2a}[/tex]
[tex]s=\frac{v^2_2-v^2_1}{2cv}[/tex]
Hence, the distance covered by the particle is [tex]s=\frac{v^2_2-v^2_1}{2cv}[/tex]
