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}