Derivation of the Equations of Motion | deriving ‘suvat equations’

Last updated on December 28th, 2022 at 08:11 am

Derivation of the Equations of Motion: To calculate quantities involving motion in a straight line at a constant acceleration we need four equations. These equations are often informally referred to as the ‘suvat equations’ after the symbols used for the 5 quantities involved.

We will derive equations for the following 5 motion quantities: displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t). Before going for the derivation, we will draw a velocity-time graph in the next section.

Velocity time graph required to derive the motion equations

This velocity time graph depicts a motion of a body in a straight line at a constant acceleration a. It has an initial velocity of u and a final velocity of v.
t is the time elapsed.
During this time duration, the displacement of the body in motion is s.

How to derive the equation for final velocity? | Final Velocity Equation without mention of displacement

From the graph above, acceleration a = gradient of velocity-time graph= Δv/Δt = (v-u)/t
=> v=u + at ……………….. (1)
This is the equation of the final velocity v.

Please note that this equation of final velocity doesn’t mention displacement s.

How to derive the equation for displacement? | Displacement Equation without mention of the final velocity

The area under the velocity-time graph is equal to the displacement s.
This v-t area in the above diagram has 2 distinct areas: one rectangular and the other one triangular.

The rectangular area of the velocity time graph = ut
And the triangular area=(1/2).(v-u).t =(1/2).(u+at-u).t=(1/2)at²

So total area under the v-t graph = displacement = s= ut + (1/2)at²
s= ut + (1/2)at² ……….(2)
This is the equation of the displacement s.

Please note that this equation of displacement doesn’t have any mention of final velocity v.

How to derive the equation for displacement? | Displacement Equation without mention of acceleration

If the area under the graph is taken as the area of a trapezium, then it has u and v as 2 parallel sides of the trapezium, and t is the perpendicular separation between these 2 sides.
So area of this trapezium=s = (1/2)(u+v)t

So displacement = average velocity * time = [(u+v)/2]. t ………..(3)

Please note that this equation of displacement doesn’t have any mention of acceleration a.

How to derive the equation for final velocity | final velocity Equation without mention of time

from equation 1 we get t=(v-u)/a

so in equation (3) replacing t we get:
s=[(u+v)/2].t = [(v+u)/2].[(v-u)/a] = (v²-u²)/(2.a)
v²-u²=2as => v²=u² + 2as………………(4)
This is another equation of the final velocity v.

Please note that this equation of final velocity doesn’t have any mention of time t.

Summary | Takeaway | Suggested reading

We have derived 4 motion equations involving 5 quantities in easy steps. As we are done with the derivation of the suvat equations, now it’s time to solve some numerical problems of the motion chapter.

Here are suggested posts for your further study and problem-solving.

Suggested reading

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