Learn the Derivation of Motion Equations with Expert Guidance

[Omar]

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..

 

[Fredrik]

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.

[tex]x''(t)=a[/tex]

[tex]x'(t)=at+v_0[/tex]

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

[tex]x(t)=\frac{1}{2}at^2+v_0t[/tex]

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.

 

[Mahfuz]

fundamentally they were derived from equation of acceleration.

:)

 

[ArmoSkater87]

start off with the definition of acceleration...
[tex]a = \frac{v_f-v_0}{t}[/tex]
[tex]v_f = v_0 + at[/tex]

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

substitute...

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

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