• question_answer A constant power P is applied to a particle of mass m. The distance traveled by the particle when its velocity increases from ${{v}_{1}}$, to ${{v}_{2}}$is (neglect friction) A) $\frac{m}{3P}(v_{2}^{3}-v_{1}^{3})$ B) $\frac{m}{3P}({{v}_{2}}-{{v}_{1}})$ C) $\frac{3p}{m}(v_{2}^{2}-v_{1}^{2})$ D) $\frac{m}{3P}(v_{2}^{2}-v_{1}^{2})$

 [a] $P=Fv=mav$ $\Rightarrow$ $a=\frac{P}{mv}$ $\Rightarrow$ $v\frac{dv}{ds}=\frac{P}{mv}$
 $\Rightarrow$ ${{v}^{2}}dv=\frac{P}{m}ds$ $\Rightarrow$ $\frac{P}{m}\int_{\,0}^{\,s}{ds}=\int_{\,{{v}_{1}}}^{\,{{v}_{2}}}{{{v}^{2}}dv}$ $\Rightarrow$ $\frac{P}{m}s=\frac{1}{3}(v_{2}^{3}-v_{1}^{3})$ $\Rightarrow$ $s=\frac{m}{3P}(v_{2}^{3}-v_{1}^{3})$