Question

A constant power $P$ is applied to a particle of mass $m$. The distance travelled by the particle when its velocity increases from $v_1\text{to}{v}_2$ is - (neglect friction)

• A. $\frac{3P}{m}(v_2^2-v_1^2)$
• B. $\frac{m}{3P}(v_2-v_1)$
• C. $\frac{m}{3P}(v_2^3-v_1^3)$
• D. $\frac{m}{3P}(v_2^2-v_1^2)$

Answer 1

The correct option is C $\frac{m}{3P}(v_2^3-v_1^3)$

We know that
Power $P=Fv$
$\Rightarrow m{v}^2\frac{dv}{dx}=P[F=ma=mv\frac{dv}{dx}]$
$\Rightarrow m{\int }_{v_1}^{v_2}{v}^{2}dv=P{\int }_0^{s}dx$
(Since P is constant)
$\Rightarrow Ps=\frac{m}{3}{\left[v^3\right]}_{v_1}^{v_2}$
$\Rightarrow s=\frac{m}{3P}[v_2^3-v_1^3]$