Final velocity of an object accelerated across a given distance with constant power

1. Aug 10, 2007

HalfThere

1. The problem statement, all variables and given/known data
An object of mass M begins with a velocity of 0 m/s at a point. A power input of P watts goes directly to kinetic energy until the object has traveled a distance of X meters. What is the final velocity of the object?

So, we have constant variables
M = mass
X = distance that power will be input
P = power level

And also
V = final velocity, after traveling distance X (to be solved)

2. Relevant equations

E = 1/2MV$$^{2}$$
X = $$\int$$V(t) dt - V(t) is V as a function of time.
E = P*t

3. The attempt at a solution

Find V as a function of E (easy)
V = $$\sqrt{2E}$$/M

Find V as a funciton of time t (use equation)
V = $$\sqrt{2P*t}$$/M

Now take the equation for X
X = $$\int$$V(t) dt

And Find X as a function of t directly, knowing the V(t) function
X = $$\int$$$$\sqrt{2P*t}$$/M dt Edit: Should be, and was calculated as ($$\int$$ $$\sqrt{2P*t}$$ dt)/M

Integrate (remember that sqrt(2P) is a constant)
X(t) = (2/3)*$$\sqrt{2P}$$*t$$^{3/2}$$/M

Now change to t in terms of X
t(X) = ((3/2)/$$\sqrt{2P}$$*M*X)$$^{2/3}$$

And finally slide that into the V(t) equation
V(X) = $$\sqrt{2P}$$*((3/2)/$$\sqrt{2P}$$*M*X)$$^{1/3}$$/M

Simplify (whew!)
V(X) = 3$$^{1/3}$$*2$$^{-1/6}$$*P$$^{1/3}$$*X$$^{1/3}$$*M$$^{-2/3}$$

So, V correlates directly with the cube root of X, the cube root of P, and M^(-2/3), with a weird constant.

Am I right? Am I not? If not, where did I go wrong?

Last edited: Aug 10, 2007
2. Aug 10, 2007

learningphysics

Your general solution is fine, but I think you made some calculation errors... your m should be under the square root... so that affects your final exponent on m... also I'm getting that the 2 cancels out (ie the power of 2 is 0)...

I'm getting: $$3^{1/3}*p^{1/3}*x^{1/3}*m^{-1/3}$$ as the final velocity. Though I may have screwed up...

Last edited: Aug 10, 2007
3. Aug 10, 2007

HalfThere

Which specific step does the miscalculation occur? I can't quite see what you mean.

Also, a more general question about energy: What will a constant power input do to something with a velocity with no immediate context, like a rocket in space?

On the one hand, I think acceleration should remain constant, because it's change in momentum is relative only to itself, but on the other hand, a constant rate of fuel consumption should mean a a constant power input. The two don't seem to be compatible. What gives?

Last edited: Aug 10, 2007
4. Aug 10, 2007

rootX

Integrate (remember that sqrt(2P) is a constant)

Now change to t in terms of X

here?

edit: the step where u isolated t.

5. Aug 10, 2007

HalfThere

I see where I went wrong with the 2^(-1/6), it should just be 3^(1/3) for the constant, but I still don't see the problem with the M. It was transformed from M to M^(2/3), then square rooted to M^(1/3), which was over M, which is M^(-2/3).

6. Aug 10, 2007

learningphysics

Here is the mistake... the M should be under the square root.

7. Aug 10, 2007

HalfThere

Ah, you're indeed right. I see now. And it makes sense if you look at units too.

Okay, Thanks learningphysics!

8. Aug 10, 2007

learningphysics

Acceleration isn't constant for constant power...

9. Aug 10, 2007

learningphysics

you're welcome.

10. Aug 10, 2007

HalfThere

Can anyone answer the third post of the topic, where I ask about conservation of momentum V. conservation of energy in the case of the rocket?