Deriving Kinematics Equations from First Principles

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Homework Help Overview

The discussion revolves around deriving kinematics equations from first principles, particularly focusing on 2D motion. Participants are exploring how to derive the equations for vertical motion using calculus, as suggested by a physics teacher.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to derive the equation for vertical velocity (Vy) from the vertical position equation (y) using calculus. There is a discussion about the process of differentiation and how it relates to the derivation of kinematics equations.

Discussion Status

Some participants have provided guidance on using calculus for the derivation, and there is an acknowledgment that this method can apply to other equations of motion in kinematics. The conversation is exploring the foundational concepts of acceleration and differentiation without reaching a consensus on all aspects of the derivation process.

Contextual Notes

Participants are working under the premise of deriving equations for an upcoming test, with a focus on understanding rather than memorization. There is an assumption that participants have some background in calculus and kinematics.

jackrc11
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Homework Statement


For an upcoming test on 2D Motion, my Physics Teacher recommended that, instead of memorizing close to a dozen equations, he suggested we derive what we can from the main equations he showed us. So people in my class decided to split up deriving equations to bring them together later for everyone to use. I was told to use these two:
y = (Vosin(theta))t - 1/2gt^2
Vy = Vosin(theta) - gt

So I was attempting to derive Vy from y, but found an issue.

Homework Equations


y = (Vosin(theta))t - 1/2gt^2
Vy = (Vosin(theta)) - gt
V = d/t

The Attempt at a Solution


To derive Vy, I simply did ((Vosin(theta))t - 1/2gt^2)) / t. This gave me (Vosin(theta)) - 1/2gt. This differs from the given equation for Vy, so I'm curious what I have to do to cancel out the 1/2.
 
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Do you know any calculus? That's the way to derive the equations.
 
axmls said:
Do you know any calculus? That's the way to derive the equations.

Yes, I am in calculus this year! I didn't even think to find d/dt. This makes things a lot easier, thanks!

Using calculus, just to be sure I'm right, you can find

y' = Vy
Vy = 1 * (Vosin(theta)) - 1/2gt * 2, giving
Vy = (Vosin(theta)) - gt

That works out really well! Can that apply to a lot of other equations of motion (in Kinematics) as well?
 
jackrc11 said:
Can that apply to a lot of other equations of motion (in Kinematics) as well?

That's where kinematics equations come from!

It all comes from the fact that a = \frac{d^2 x}{dt^2} and the assumption that a is constant.

If you've learned about antidifferentiation (essentially the opposite of differentiation), then you can derive your equations from that equation alone. Since a = \frac{d v}{dt}, v = \int a \ dt = v_0 + a t. And since v = \frac{d x}{dt}, x = \int v \ dt = x_0 + v_0 t + \frac{1}{2} a t^2. If you haven't learned about antidifferentiation, you can at least confirm that you can work backwards from those by taking derivatives to get back to \frac{d^2 x}{dt^2} = a.

The equation that is not time-dependent is a bit trickier to derive, but that's essentially how the "usual" kinematics equations are derived. It all comes back to calculus--the mathematics of change.
 
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