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

Last updated on May 3rd, 2023 at 05:39 pm

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.

- Derivation of the Equations of Motion (Suvat equations)
- Velocity time graph required to derive the motion equations
- Derivation of final velocity equation (without mentioning displacement)
- Derivation of Displacement Equation (without mentioning final velocity)
- Derivation of Displacement Equation (without mentioning acceleration)
- Derivation of final velocity Equation (without mention of time)

## Derivation of the Equations of Motion (Suvat equations)

Now, let us start with the derivation of suvat equations or motion equations. First, we will draw a velocity-time graph, and then using that we will derive the equations of motion one by one.

### 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.

### Derivation of final velocity equation (without mentioning displacement)

From the graph above, acceleration **a** = gradient of velocity-time graph= * Δ*v/

*t = (v-u)/t*

**Δ****=> 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.**

### Derivation of Displacement Equation (without mentioning 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)at^{2}

So total area under the v-t graph = displacement = s= ut + (1/2)at^{2}**s= ut + (1/2)at ^{2} ……….(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.**

### Derivation of Displacement Equation (without mentioning 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.**

### Derivation of 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] = (v^{2}-u^{2})/(2.a)**v ^{2}-u^{2}=2as => v^{2}=u^{2 }+ 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.