# Power and velocity

1. Apr 22, 2015

### elitewarr

1. The problem statement, all variables and given/known data
The rocket sled has a mass of 4 Mg and travels along the smooth horizontal track such that it maintains a constant power output of 450 kW. Neglect the loss of fuel mass and air resistance, and determine how far the sled must travel to reach a speed of v = 60 m>s starting from rest.

2. Relevant equations

P = Fv
$E_k = \frac{1}{2} mv^2$
W=Fs

3. The attempt at a solution

Since the rocket sled is moving along a flat surface, and that there is no loss of energy, we can conclude that the energy coming from the power output will be converted to kinetic energy.

$E_k = W = Fs = \frac{P}{v} s$

Thus simplifying, we get

$\frac{1}{2} mv^3 = Ps$

But apparently, there is something wrong with the equations since if we go about from another way,

$P = Fv = (ma)v = (mv\frac{dv}{ds})v$

$\int_0^s P \, ds = m\int_0^v v^2 \, dv$

$Ps = \frac{1}{3} mv^3$

Is there something wrong with my concept or that I am not seeing?

Thank you.

2. Apr 22, 2015

### haruspex

Force will not be constant here, so that needs to be an integral, not a simple product.

3. Apr 22, 2015

### elitewarr

Thank you for the fast reply.

Okay. I was suspecting much. But how do I do the integration for the $\frac{1}{v}$ with respect to ds? Any clues as to where to start?

Last edited by a moderator: Apr 22, 2015
4. Apr 22, 2015

### Staff: Mentor

might be useful: Energy = power x time

5. Apr 22, 2015

### HallsofIvy

Staff Emeritus
Since v= ds/dt, $\int \frac{P}{v}ds= P\int \d Since [itex]v= \frac{ds}{dt}$,
$$P\int \frac{1}{v}ds= P\int \frac{1}{\frac{ds}{dt}}ds= P\int \frac{dt}{ds}ds= P\int dt= P(t_1- t_0)$$

6. Apr 22, 2015

### haruspex

Relate F to dv/dt.
Edit: or better still, to vdv/ds.

Last edited: Apr 22, 2015