# Kinematics particle velocity problem

1. Sep 13, 2008

### musicfairy

1. The problem statement, all variables and given/known data

This is from the 1983 ap test. I'm using it to study the concepts but so far it only confused me more.

It goes:

A particle moves so that the x-component of its velocity has the constant value vx = C; that is, x = Ct

1. Determine the y-component of the particle's velocity as a function of x.

2. Determine the y-cpmponent of the particle's acceleration.

Part b.

Suppose, instead, that the particle moves along the same parabola with a velocity whose x-component is given by vx = C/(a+x2)1/2

3. Show that the particle's speed is constant in this case.

3. The attempt at a solution

I have the solution, but it doesn't make any sense to me. For the first question they showed dy/dt = (dy/dx)(dx/dt) They said it's the chain rule, but where did that come from? I thought I knew the chain rule until I saw that. Where did t come from?

I can't figure out where the answer to the 2nd question is from either. They put ay = (dvx/dt) = (d/dt)(C2t) = C2
I can't figure out where that came from either.

Number 3 confuses me even more than the previous 2. This is what they did:

v = sqrt(vx2 + vy2)

vy = dx/dt = (dy/dx)(dx/dt) = xvx

v = sqrt((vx2)(1 + x2)) = sqrt( (C2/ 1+x2)(1 + x2)) = C

If you have access to 1983 mech #1 everything looks much better than what I typed.

Can someone please explain all this to me? I'm trying to prepare for the ap test by looking at old problems, but so this one is written in hieroglyphics and I could really use some help.

And please explain how the chain rule makes dy/dx = (dy/dx)(dx/dt)

Any help is greatly appreciated.

Last edited: Sep 13, 2008
2. Sep 13, 2008

### Clairefucious

$$\frac{dy}{dt}$$ may be written as $$\frac{dy}{dt}\frac{dx}{dx}$$. Rearranging we get $$\frac{dy}{dx}\frac{dx}{dt}$$. Does this make a bit more sense by writing in the middle step?

For acceleration $$a = \frac{dv}{dt}$$ and velocity is $$v \frac{dx}{dt}$$. We put these together by substituting v to get $$a = \frac{d2x}{dt2}$$.

3. Sep 13, 2008

### Clairefucious

That last formula was meant to be

$$a = \frac{d^{2}x}{dt^{2}}$$

4. Sep 13, 2008

### musicfairy

Thanks for the chain rule explanation. It makes a lot more sense now.