Kinematics particle velocity problem

musicfairy
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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 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.



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:
Comments on the chain rule:

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

For acceleration [tex]a = \frac{dv}{dt}[/tex] and velocity is [tex]v \frac{dx}{dt}[/tex]. We put these together by substituting v to get [tex]a = \frac{d<sup>2</sup>x}{dt<sup>2</sup>}[/tex].
 
Clairefucious said:
Comments on the chain rule:

That last formula was meant to be

[tex]a = \frac{d^{2}x}{dt^{2}}[/tex]
 
Thanks for the chain rule explanation. It makes a lot more sense now.
 

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