Calculating Velocity and Acceleration Along a Parabolic Path

[glid02]

Hey guys,

Here's the question:

A point A moves along a curve with the equation x = y^2/6 and is elevated in the y direction at a constant velocity of 3 in/s. Calculate the velocity and the acceleration when x = 6 in.

So I solved for y and y=sqrt(6*x) and when x=6 y=6.

Since the point is moving at a constant velocity of 3 in/s it takes 2 seconds to get to 6 in.

I plugged this into y=y0+vt+1/2at^2 and got a=0 which is obvious since it's moving at a constant velocity.

Here's where I'm stuck, I tried the same equation with x. I think I may be able to solve for v and a by solving the differential equation but I don't think that's what I'm supposed to do.

It seems like I'm supposed to differentiate the positions with the equation giving but I'm not sure how I'm supposed to do that because when differentiating y with respect to x the derivative of y=sqrt(6*x) is 0 and the derivative of x=y^2/6 is also 0.

Any help would be great. Thanks a lot.

 

[chanvincent]

You have a very important piece of information: dy/dt is constant...
and you are asked to find the velocity and accelaration when x = 6...
since dy/dt and d^2y/dt^2 is known( how do you get the accelaration in the y direction?)
, all you need is dx/dt, and d^2x/dt^2, right?

differentiate x=y^2/6 over t on both side... you will have dx/dt in terms of dy/dt and y...
differentiale it once more, you will get d^2x/dt^2, which is a constant...

EDIT: One more hint.. Use chain rule: $$\frac{dy^2}{dt} = \frac{dy^2}{dy}\frac{dy}{dt} = 2y\frac{dy}{dt}$$

 

[m2003]

Hi there,

Thank you for sharing your question with us. It seems like you have made some good progress in solving the problem so far. I can see that you have correctly solved for y and found that when x = 6, y = 6. Additionally, you have correctly calculated that the point takes 2 seconds to reach x = 6 at a constant velocity of 3 in/s. This means that the velocity at x = 6 is also 3 in/s.

Now, to find the acceleration, we need to differentiate the velocity equation with respect to time. Since we know that the velocity is constant (3 in/s), the derivative will be 0, as you have correctly pointed out. This means that the acceleration is also 0 at x = 6.

It is important to remember that in this problem, the velocity and acceleration are in the y direction, not the x direction. So, even though the point is moving along a curve in the x direction, its velocity and acceleration are only in the y direction. This is why differentiating the position equation with respect to x is not necessary.

I hope this helps clarify things for you. Keep up the good work!