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Homework Help: Physics - Calculus question - Convservation of energy

  1. Aug 2, 2010 #1
    Ok here is my question:

    Potential energy = mgh
    Kinetic energy = 1/2mv^2
    mgh = 1/2mv^2, solving for v you get v = sqrt(2gh)

    Now I know v is the same as dx/dt so if I substite in dx/dt for v :

    dx/dt = sqrt(2gh)

    Multiply by dt:
    dx = sqrt(2gh) * dt
    x = t*sqrt(2gh)

    This should be right, but for some reason it doesnt work when I check it with the well known x = x0 + v0t + 1/2at^2 formula, what am I doing wrong?
  2. jcsd
  3. Aug 2, 2010 #2
    v is the velocity at the end of the free fall.It is determined by the initial height and it is equal to sqrt(2gh) it is constant. In order to substitute the derivative dy/dt it must refer to the rate of change of height with respect to time from the start to the finish of the fall.
  4. Aug 2, 2010 #3
    no no no no, well everything the fella above me said is true, but I assume you already accounted for t being the end of the fell from start to finish

    that is not the problem. dx/dt equels AVERAGE velocity and not the final velocity. quite two different things.

    assuming constant acceleration if you do it by average velocity * 2 = final velocity, and hence 2x = t * sqrt(2gh) you'll find everything to be correct.

    if acceleration isn't constant there will be no way to work in the average velocity.
  5. Aug 3, 2010 #4
    Thanks for your help guys, the problem was that h was a variable and dependent on time, and not a constant.
  6. Aug 3, 2010 #5
    you are wrong my friend, that was not your problem.

    Vfinal = dx/dt is not a statement you can make. Vfinal = dx/dt only when a = 0 which is clearly not the case in a free fall.

    take for example two situations:

    a car is cruising at a constant speed over the period of 3 sec it passes 45m, how fast was it going after 3 seconds? here you can apply v = dx/dt = 15m/sec;

    now a car is accelerating from rest at a constant speed over a period of 3 seconds during that period it passes 45m, how fast is it going after 3 seconds? 15m/sec? that's obviously not true.

    if you'll solve
    45 = 0.5*a*3^2... a = 10;
    v = 0 + a*t = 10*3 = 30. in the first case Vfinal = 15, second case Vfinal = 30; the average v which is what you get in Vavg = dx/dt is 15m/s in BOTH.
  7. Aug 3, 2010 #6


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    No, PhysicsPrac was right. One way to do the problem correctly would be to treat h as a variable.
    It sounds like you're misunderstanding the meaning of dx/dt. It refers to instantaneous velocity, i.e. the velocity at a specific instant, but you are talking about it as if it were average velocity, the net displacement per unit time over a time interval.
    You are correct that vavg = 15 m/s in both cases, but it is not true that dx/dt = 15 m/s in the second case. dx/dt changes with time; its value increases from 0 m/s to 30 m/s over the 3-second interval.

    It would be correct to say that Δx/Δt = 15 m/s in the second case; maybe that's what you meant.
  8. Aug 3, 2010 #7
    I thought d means delta or Δ.
    Since you say it isn't and since you agree that the instantaneous velocity at t = 3 is 30m/s. explain to me with what values will dx/dt will give you 30m/s.

    and I don't see how "treating h as a variable" helps, ofcourse it's a variable and given the value h = x his statement is still false and the two equations x = x0 + v0*t + 0.5at^2 and x = t*sqrt(2gh) will still give different results.
    Last edited: Aug 3, 2010
  9. Aug 3, 2010 #8
    on top of that as far as I understand it now dx/dt is simply a notation saying the derivative of x as a function of t.

    hence I don't see how
    is legal.

    Infact dx/dt is equel to v0 + a*t since it's the derivative of x0 + v0*t + 0.5at^2 and if you substitute that for "v" it solves fine.
  10. Aug 3, 2010 #9
    Messing around with the differentials is legal(in Physics:rofl:).The problem is that you must be very careful when you do it. In this case it does not work if you don't have h as a function of t. And if you have it why do you need to calculate this anyway?
  11. Aug 3, 2010 #10


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    Nope, it doesn't. Δ represents the change in a quantity, its final value minus its initial value, but the d in a derivative is something more complicated. It doesn't really mean anything by itself, it's only the whole expression dx/dt that has a meaning, and that meaning is "the derivative of the function x(t)".
    Good call noticing that. That's an abuse of notation - or, if you ask a mathematician, cheating ;-) It's a technique called separation of variables, and although it seems wrong, it does actually work; you just have to be careful about what is constant and what isn't.
    Values of what? It's not like dx is a variable that has some particular numeric value. Like I said above, dx/dt is a notation for the derivative of x(t), and you can't sensibly break it into parts.
    Well, the important thing is that h is time-dependent. That makes a difference when you do the integration. Everything that PhysicsPrac did up to
    [tex]\mathrm{d}x = \sqrt{2gh}\mathrm{d}t[/tex]
    is technically correct (though the notation is a bit confusing), but to integrate it, you can't just pull [tex]\sqrt{2gh}[/tex] out of the integral, because it's not a constant. h represents the height of the object, and that height changes with time.

    Note that velocity is the rate of change in position, and in this case, the position is height. If you use h to represent height, then v should be the derivative of h. It's confusing to write v = dx/dt when you already have a variable to represent position. Isiudor, you might want to try going through this with v = dh/dt instead of v = dx/dt and see how it works out.
    Last edited: Aug 3, 2010
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