Kinematics particle velocity problem

AI Thread Summary
The discussion revolves around a particle velocity problem from a 1983 AP test, focusing on determining the y-component of velocity and acceleration based on given x-velocity equations. Participants express confusion over the application of the chain rule in deriving dy/dt and the acceleration formula, seeking clarification on the steps involved. The chain rule is explained, emphasizing the relationship between dy/dt, dy/dx, and dx/dt, which helps clarify the derivation process. Additionally, the constant speed of the particle in part three is discussed, with participants working through the velocity equations to understand the underlying concepts. Overall, the thread highlights the challenges faced by students in grasping kinematic principles and the importance of clear explanations in mastering these concepts.
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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:
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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 a = \frac{d<sup>2</sup>x}{dt<sup>2</sup>}.
 
Clairefucious said:
Comments on the chain rule:

That last formula was meant to be

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