Power and velocity as a function of time

AI Thread Summary
The discussion revolves around a physics problem involving a steam engine with a constant power output of 1.5 MW, which accelerates from 10 m/s to 25 m/s in 6 seconds. The first part of the problem requires calculating the mass of the steam engine using the relationship between power, work, and velocity changes. The second part involves deriving the velocity as a function of time, leading to a differential equation that needs to be solved. A key point raised is the importance of correctly incorporating initial conditions into the solution, particularly the initial velocity of 10 m/s, which was initially overlooked. Clarifications on integrating the differential equation with proper limits help resolve confusion regarding the integration constant.
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Homework Statement



A steam-engine is traveling along a rail, with a constant power output of 1.5MW, regardless of its velocity.

1. What is the mass of the steam-engine if it is known that it accelerates from 10\frac{m}{s} to 25\frac{m}{s} in 6 sec?
2. Describe the velocity of the steam-engine as a function of time.
And the other questions follow from these two.

Homework Equations


P=\vec F \cdot \vec V=\frac{\Delta W}{\Delta t}=constant

The Attempt at a Solution


1.

P=\frac{\Delta W}{\Delta t}=\tfrac{1}{2}\frac{(v_f^2-v_0^2)}{\Delta t}m

m=\frac{2P\Delta t}{v_f^2-v_0^2}
//

2.

F=\frac{P}{v}

a=\tfrac{P}{m}v^{-1}

\dot v=\tfrac{P}{m}v^{-1} Here we have a simple differential equation.

\frac{dv}{dt}=\tfrac{P}{m}\frac{dt}{dx}

\frac{dv}{dx}\frac{dx}{dt}=\tfrac{P}{m}\frac{dt}{dx}

v\cdot dv=\tfrac{P}{m}dt

\tfrac{1}{2}v^2=\tfrac{P}{m}\cdot t+C

Now how do I get from here to the final solution? Omitting the +C completely provides a function v(t) that is indeed a solution to the differential equation. However, it assumes that v_0=0
I can't quite find a way to incorporate the initial values into the solution.

Well, I did find a way, but it doesn't quite make sense mathematically, I'd love a thorough explanation as to why the following is correct:

\tfrac{P}{m}\equiv \tau
v(t)=\sqrt{2\tau\cdot t+v_0^2}

Taking the time-derivative provides us with:
\dot v(t)=\frac{1}{2\sqrt{2\tau\cdot t+v_0^2}}\cdot 2\tau
The above must also be equal to: \tau v^{-1} in order to satisfy the differential equation.
\tau v^{-1}=\tau \frac{1}{\sqrt{2\tau\cdot t+v_0^2}}

So the differential equation is satisfied.

Checking that against the data in question #1, shows that it holds, and it makes sense, since at t=0\rightarrow v=v_0, however, it seems as though we used the integration constant quite haphazardly, so I'd love an explanation. :)

Is it just a matter of isolation v(t)=\sqrt{2\tau\cdot t+2C}

And then finding, using the initial values of the problem that 2C=v_0^2 ?With thanks in advance, Anatoli.
 
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The problem asks you to find the velocity as a function of time. Assume that that "time" starts (t = 0) when the steam-engine is traveling at 10 m/s. Put that in your equation and solve for C. It would be clearer to you what is going on if you had limits to your integration of the diff. eq., namely v0 to v on the left and t0 to t on the right.
 
kuruman said:
The problem asks you to find the velocity as a function of time. Assume that that "time" starts (t = 0) when the steam-engine is traveling at 10 m/s. Put that in your equation and solve for C. It would be clearer to you what is going on if you had limits to your integration of the diff. eq., namely v0 to v on the left and t0 to t on the right.

Oof! Big mistake on my part, thanks a lot for helping clear that up. :)
I assumed v_0=0 so no wonder that showed up in my result!
 
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