- #1

jeebs

- 325

- 4

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 (x

_{A},y

_{A}) to the lower position B (x

_{B},y

_{B}). 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 (x

_{A},y

_{A}) and (x

_{B},y

_{B}).

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, v

^{2}= u

^{2}+ 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 v

_{i}/v = 0 at time t=0 and position A, and increases to v

_{i}/v = 1 at time t=t

_{tot}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.