Kinematics particle velocity problem

[musicfairy]

Homework Statement

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 v_x = 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.

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 a_y = (dv_x/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(v_x² + v_y²)

v_y = dx/dt = (dy/dx)(dx/dt) = xv_x

v = sqrt((v_x²)(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.

 

[Clairefucious]

Comments on the chain rule:

$$\frac{dy}{dt}$$ 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 [tex] a = \frac{d²x}{dt²}[/tex].

 

[Clairefucious]

Comments on the chain rule:

That last formula was meant to be

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

 

[musicfairy]

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