The total travel time along a frictionless path?

[jeebs]

Hi,
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:

$$t_t_o_t = \frac{1}{\sqrt{2g}}\int dt \sqrt{\frac{(dx/dt)^2 + (dy/dt)^2}{y}}$$
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, 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, $$v = \sqrt{2gy}$$.

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

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

$$(\sqrt{(dx/dt)^2 + (dy/dt)^2})dt = \sqrt{dt^2(dx/dt)^2 + dt^2(dy/dt)^2} = \sqrt{(dx)^2 + (dy)^2} = dy\sqrt{(dx/dy) + 1}$$

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.

 

[chrisk]

What is the direction of the instantaneous velocity

$$v_i$$

and compare with your final velocity direction

$$v$$

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

$$\frac{1}{\sqrt{2g}}$$

You should find the units of the expression is time.

 

[Cyosis]

Start out with ds=vdt. From this it follows that $t=\int \frac{ds}{v}$. You know that $ds=\sqrt{dx^2+dy^2}$ 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.

 

[jeebs]

Start out with ds=vdt. From this it follows that $t=\int \frac{ds}{v}$. You know that $ds=\sqrt{dx^2+dy^2}$ 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.

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 E_k = 0.5mv² and potential energy E_p = -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. E_k + E_p = 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=t_tot.

Am I justified in saying this?

 

[paranormal]

I would approach this problem by first understanding the physical principles at play. In this case, we have an object moving along a track due to a uniform gravitational field, with no friction. This means that the only forces acting on the object are gravity and normal force, which is perpendicular to the surface of the track.

To solve for the total travel time, we need to consider the object's motion in both the x and y directions. In the x direction, the object is not accelerating, so the velocity remains constant. In the y direction, the object is accelerating due to gravity, and its velocity changes over time.

Using the equations of motion, we can express the velocity in the y direction as v_y = v_i + at, where v_i is the initial velocity and a is the acceleration due to gravity. Since we know that v_i = 0 and a = g, we can simplify this to v_y = gt.

Now, to find the total travel time, we need to integrate this velocity over time. This is where the limits of the integral come into play. The limits are given as positions (xA,yA) and (xB,yB) because we are considering the object's motion from point A to point B. We can use the equation for displacement, s = ut + 1/2at^2, to express the y position as a function of time.

Substituting v_y = gt into this equation, we get y = 1/2gt^2. This allows us to express the time t in terms of y, which we can then substitute into our integral.

The integral now becomes t_t_o_t = \frac{1}{\sqrt{2g}}\int dt \sqrt{\frac{(dx/dt)^2 + (dy/dt)^2}{y}} = \frac{1}{\sqrt{2g}}\int \frac{1}{\sqrt{g}} \sqrt{dx^2 + dy^2} = \frac{1}{g}\int \sqrt{dx^2 + dy^2}

This integral represents the total distance traveled by the object from A to B, and dividing it by the constant acceleration g gives us the total time.

In summary, the limits of the integral represent the start and end points of the object's motion, and integrating the velocity over time gives us the total distance traveled. I would