Finding the distance travelled during change in velocity from $v_i$ to $v_f$

Question

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)

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

where

P=>Power

F=>Force

m=>Mass

S=>Distance

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

Very related to this SE question: http://physics.stackexchange.com/questions/211078/terminal-velocity-and-distance-traveled-with-constant-but-limited-power/211109#211109 — Gert, Oct 21 '15 at 13:56
Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. — John Rennie, Oct 21 '15 at 14:05

score 2 · Accepted Answer · answered Oct 21 '15 at 14:05

2

HINT:
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,

answered Oct 21 '15 at 14:05

Sathyaram

392

Finding the distance travelled during change in velocity from $v_i$ to $v_f$

1 Answers1