Proving a known function of position via Chain Rule

AI Thread Summary
The discussion revolves around using the Chain Rule to prove that velocity can be expressed as a function of position when acceleration is known. The integral equation provided relates the change in velocity to the integral of acceleration over position. Participants express confusion regarding the correct application of the Chain Rule and the integral setup, with one noting a mistake in the equation's sign. A hint is given to consider the left-hand side as an integral of velocity with respect to velocity. The conversation highlights the importance of correctly applying mathematical principles to derive the desired relationship in rectilinear motion.
kylera
Messages
40
Reaction score
0

Homework Statement


Use the Chain Rule to prove that for rectilinear motion, when the acceleration is a known function of position, you can find the velocity as a function of position via the integral

\frac{v^{2}-v_{0}^{2}}{2} = \int^{s}_{s_{0}}a(s)ds


Homework Equations


v^{2}=v_{0}^{2}\times2as


The Attempt at a Solution


I took the left fraction, substituted v^2, simplified and got as. I let A be as, then took dA to get a da. Now I'm stuck.
 
Physics news on Phys.org
I think you simplified wrong. It's a multipication sign not addition. See what I mean?
 
Hold up, I wrote the question on the board wrong -- it is supposed to be a plus for the relevant equation part.
 
I'm not exactly sure what they're asking here. For constant accelerations, you "relevant equation" is basically the answer, assuming you swap out the "x" for a "+" and take a square root. What's throwing me is the request for proof by chain rule.
 
Chain Rule

kylera said:

Homework Statement


Use the Chain Rule to prove that for rectilinear motion, when the acceleration is a known function of position, you can find the velocity as a function of position via the integral

\frac{v^{2}-v_{0}^{2}}{2} = \int^{s}_{s_{0}}a(s)ds


Homework Equations


v^{2}=v_{0}^{2}\times2as


The Attempt at a Solution


I took the left fraction, substituted v^2, simplified and got as. I let A be as, then took dA to get a da. Now I'm stuck.

Hi kylera! :smile:

You were asked to use the Chain Rule. So …

Hint: the LHS is ∫vdv. So use the Chain Rule on dv. :smile:
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top