Power & Velocity Homework Solution | 450 kW

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The discussion focuses on calculating the distance a rocket sled must travel to reach a speed of 60 m/s, given a constant power output of 450 kW and a mass of 4 Mg. The initial approach involves using the relationship between power, force, and velocity, but the participants identify that force is not constant, necessitating the use of integration. They explore integrating the power with respect to distance and time, emphasizing the need to relate force to acceleration and velocity changes. The conversation highlights the importance of correctly applying calculus to solve the problem. Ultimately, the participants are seeking clarity on the integration process to arrive at the correct solution.
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Homework Statement


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.

Homework Equations



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

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.
 
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elitewarr said:
W=Fs
Force will not be constant here, so that needs to be an integral, not a simple product.
 
haruspex said:
Force will not be constant here, so that needs to be an integral, not a simple product.

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?
 
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might be useful: Energy = power x time
 
elitewarr said:
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?
Since v= ds/dt, \int \frac{P}{v}ds= P\int \d<br /> <blockquote data-attributes="member: 198689" data-quote="elitewarr" data-source="post: 5084413" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> elitewarr said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Thank you for the fast reply.<br /> <br /> 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? </div> </div> </blockquote> Since v= \frac{ds}{dt},<br /> 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)
 
elitewarr said:
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?
Relate F to dv/dt.
Edit: or better still, to vdv/ds.
 
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