# 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?