1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: The total travel time along a frictionless path?

  1. Apr 2, 2010 #1
    I have this problem about an object moving along a track due to a uniform gravitational field. It moves from being at rest in the higher position A (xA,yA) to the lower position B (xB,yB). There is no friction at all. What I am supposed to be doing is showing that the total time from A to B is given by:

    [tex] t_t_o_t = \frac{1}{\sqrt{2g}}\int dt \sqrt{\frac{(dx/dt)^2 + (dy/dt)^2}{y}}[/tex]
    where the limits of the integral are given as between (xA,yA) and (xB,yB).

    The limits I do not quite understand, because surely they should really be times rather than positions if we are integrating with respect to time?

    Anyway, what I have attempted is this:
    From Newton's equation of motion, v2 = u2 + 2as, if we set initial velocity u = 0, gravitational acceleration a = g and vertical displacement s = y, then we have an expression for the object's final velocity, [tex]v = \sqrt{2gy}[/tex].

    Also, I said that the instantaneous velocity will be given by [tex] v_i = \sqrt{(dx/dt)^2 + (dy/dt)^2} [/tex].

    Since these two things match the denominator and numerator in the integrand respectively, they must be important. Clearly vi/v = 0 at time t=0 and position A, and increases to vi/v = 1 at time t=ttot and position B.

    This is where I start to hit trouble though. I'm not really sure what to make of these spatial limits and I can't see the way forward. One thing I have tried is to say that

    [tex] (\sqrt{(dx/dt)^2 + (dy/dt)^2})dt = \sqrt{dt^2(dx/dt)^2 + dt^2(dy/dt)^2} = \sqrt{(dx)^2 + (dy)^2} = dy\sqrt{(dx/dy) + 1}[/tex]

    but when I proceed with the integral integrating with respect to y from there, I end up with a quantity that does not have dimensions of time.

    Can anyone steer me in the right direction?
    Many thanks to those who try.
  2. jcsd
  3. Apr 2, 2010 #2
    What is the direction of the instantaneous velocity


    and compare with your final velocity direction


    If the dt outside the square root is moved inside the square root, as you have shown, what are the units of the integrand? Then determine the units of


    You should find the units of the expression is time.
  4. Apr 4, 2010 #3


    User Avatar
    Homework Helper

    Start out with ds=vdt. From this it follows that [itex]t=\int \frac{ds}{v} [/itex]. You know that [itex]ds=\sqrt{dx^2+dy^2}[/itex] which you can rewrite to something in terms of dt as you have already shown in your last line of equations. Now you just have to calculate v. Hint: use conservation of energy.
  5. Apr 6, 2010 #4
    Thanks guys.

    What I've tried now is this:-
    In the equation ds = vdt, this velocity is not a final or initial velocity as are used in Newton's equations of motion, it is some instantaneous velocity v=v(t) = v(y).

    Also, kinetic energy Ek = 0.5mv2 and potential energy Ep = -mgy.

    Potential energy is at a maximum of zero when the object is at its highest, and as it falls along its path it becomes more negative. At t=0, velocity is also 0, so that the sum of the kinetic and potential energies is zero, ie. Ek + Ep = 0, which is a constant.

    This means that at some downwards displacement y, we have 0.5mv2 - mgy = 0 at all times. This leads directly to v=(2gy)1/2, which then slots into my final expression for t=ttot.

    Am I justified in saying this?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook