A constant power is applied to a particle of mass m. The velocity of the particle increases from $v_i$ to $v_f$. We have to find the distance travelled by the particle during this interval.(neglect friction)

$$P=Fv_f$$ $$P=mav_f$$ $$P=\frac{mv_f(v_f^2-v_i^2)}{2S}$$ $$S=\frac{mv_f(v_f^2-v_i^2)}{2P}$$






However the correct answer is supposed to be $\frac{m(v_f^3-v_i^3)}{3P}$. Can anyone explain this.


closed as off-topic by ACuriousMind, John Rennie, Sebastian Riese, HDE 226868, user10851 Oct 22 '15 at 3:01

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Using Chain rule, We get $a= \frac{vdv}{dx}$,
$$ P=constant\\ P=F.v\\ P=(ma).v\\ \\ P=\left(m\frac{v\space dv}{dx}\right).v $$ Now Integrate,


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