Question

A particle moving in a straight line is acted by a force, which works at a constant power and changes its velocity from $u$ to $v$ in passing over a distance $x$ . The time taken will be:

• A. $x\left(\frac{v - u}{v^{2} + u^{2}}\right)$
• B. $x\left(\frac{v + u}{v^{2} + u^{2}}\right)$
• C. $\frac{3}{2}x\left(\frac{v^{2} - u^{2}}{v^{3} - u^{3}}\right)$
• D. $x\left(\frac{v}{u}\right)$

Answer 1

Answer: (C)

The force acting on the particle $=\frac{m d v}{d t}$
Power of the force $=\left(\frac{m d v}{d t}\right)v=k$ (constant)
$m\frac{v^{2}}{2}=kt+c...\left(1\right)$
at $t=0,v=u$
$\therefore c=\frac{m u^{2}}{2}$
Now from (1),
$m\frac{v^{2}}{2}=kt+\frac{m u^{2}}{2}$
$\frac{1}{2}m\left(v^{2} - u^{2}\right)=kt...\left(2\right)$
Again $\frac{m d v}{d t}v=k$
$m.v\frac{d v}{d x}v=k$
$\displaystyle \int _{0}^{v}mv^{2}dv=\displaystyle \int _{0}^{x}kdx$
Integrating, $\frac{1}{3}m\left(v^{3} - u^{3}\right)=kx\ldots \left(3\right)$
From eqn $\left(2\right)$ and $\left(3\right)$ ,
$t=\frac{3}{2}\left(\frac{v^{2} - u^{2}}{v^{3} - u^{3}}\right)x$