Power and velocity as a function of time

[RoyalCat]

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.

 

[kuruman]

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.

 

[RoyalCat]

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!