Learn the Derivation of Motion Equations with Expert Guidance

  • Context: High School 
  • Thread starter Thread starter Omar
  • Start date Start date
  • Tags Tags
    Derivation
Click For Summary

Discussion Overview

The discussion revolves around the derivation of motion equations, specifically s=ut+0.5at^2 and v^2=u^2+2as. Participants explore the theoretical foundations and mathematical steps involved in deriving these equations, focusing on concepts such as acceleration, velocity, and position as functions of time.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant states that the motion equations are derived from the definition of acceleration as the second derivative of position when acceleration is constant, involving integration to find velocity and position.
  • Another participant emphasizes that the equations fundamentally arise from the definition of acceleration.
  • A different participant provides a step-by-step derivation starting from the definition of acceleration and average velocity, leading to the first motion equation.
  • This participant also mentions that the second equation can be derived through manipulation of the first equation and substitution of acceleration definitions.

Areas of Agreement / Disagreement

Participants present various derivations and approaches, but there is no consensus on a single method or perspective. Multiple competing views on the derivation process remain evident.

Contextual Notes

The discussion includes various mathematical manipulations and assumptions about initial conditions, such as the choice of position at time t=0, which are not universally agreed upon.

Omar
Messages
13
Reaction score
0
Hello,

I'm new to the forums. This is my first thread here. I just wanted help in knowing where the motion equations (s=ut+0.5at^2, and v^2=u^2+2as) are derived from?

Thanks in advance..
 
Physics news on Phys.org
They are a consequence of the definition of acceleration as the second derivative of position (as a function of time). When the acceleration is constant, it's easy to find the velocity by doing a very simple integration.

x''(t)=a

x'(t)=at+v_0

The constant term is the velocity at time t=0. If we integrate this, we find the position as a function of time.

x(t)=\frac{1}{2}at^2+v_0t

I could have added a new constant term that would have represented the position at time t=0, but this is usually chosen to be zero.

The second equation in your post can be derived from the velocity and position equations above.
 
fundamentally they were derived from equation of acceleration.

:)
 
start off with the definition of acceleration...
a = \frac{v_f-v_0}{t}
v_f = v_0 + at

Definition of average velocity...
v_{ave} = d/t
d = \frac{v_f + v_0}{2}t

substitute...

d = \frac{(v_0+at) + v_0}{2}t
d = \frac{2v_0 + at}{2}t
d = v_0t + \frac{1}{2}at^2

For the second equation...it just a lot of manipulation of the equation above and a substitution of the definition of acceleration.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 49 ·
2
Replies
49
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 13 ·
Replies
13
Views
104K
  • · Replies 15 ·
Replies
15
Views
1K