Calculating Final Velocity of an Object with Power Input

[HalfThere]

Homework Statement

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)

Homework Equations

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

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 = \intV(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?

 

[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...

 

[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?

 

[rootX]

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

Integrate (remember that sqrt(2P) is a constant)
X(t) = (2/3)*LaTeX graphic is being generated. Reload this page in a moment.*tLaTeX graphic is being generated. Reload this page in a moment./M

Now change to t in terms of X
t(X) = ((3/2)/LaTeX graphic is being generated. Reload this page in a moment.*M*X)LaTeX graphic is being generated. Reload this page in a moment.here?

edit: the step where u isolated t.

 

[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).

 

[learningphysics]

V = \sqrt{2E}/M

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

 

[HalfThere]

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

Okay, Thanks learningphysics!

 

[learningphysics]

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?

Acceleration isn't constant for constant power...

 

[learningphysics]

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

Okay, Thanks learningphysics!

you're welcome.

 

[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?