Deriving Kinematics Equations from First Principles

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SUMMARY

The discussion focuses on deriving kinematics equations from first principles using calculus. Participants explored the equations for vertical motion, specifically y = (Vosin(theta))t - 1/2gt^2 and Vy = Vosin(theta) - gt. By applying differentiation, they confirmed that Vy can be derived as the derivative of y, leading to the conclusion that kinematics equations fundamentally stem from the principles of calculus, particularly through differentiation and antidifferentiation.

PREREQUISITES
  • Understanding of basic calculus concepts, including differentiation and antidifferentiation.
  • Familiarity with kinematics equations in physics.
  • Knowledge of the relationship between acceleration, velocity, and displacement.
  • Ability to manipulate trigonometric functions in the context of motion.
NEXT STEPS
  • Study the derivation of kinematics equations using calculus techniques.
  • Learn about the applications of differentiation in physics, particularly in motion analysis.
  • Explore the concept of constant acceleration and its implications in kinematics.
  • Investigate the relationship between velocity and displacement through integration.
USEFUL FOR

Students preparing for physics exams, educators teaching kinematics, and anyone interested in the mathematical foundations of motion in physics.

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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