How to Derive of the Equations of Motion easily? | derivation ‘suvat’

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 u and final velocity v. t is the time elapsed. During this time duration, the displacement of the body in motion is

Read the complete post with diagram and derivation on my Physics Blog: Derivation of motion equations (SUVAT)

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=

Read the complete post with diagram and derivation on my Physics Blog: Derivation of motion equations (SUVAT)

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.

Read the complete post with diagram and derivation on my Physics Blog: Derivation of motion equations (SUVAT)

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…

Read the complete post with diagram and derivation on my Physics Blog: Derivation of motion equations (SUVAT)

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] = (v 2-u 2)/(2.a)

Read the complete post with diagram and derivation on my Physics Blog: Derivation of motion equations (SUVAT)

Summary | Takeaway | Suggested reading

Read the complete post with diagram and derivation on my Physics Blog: Derivation of motion equations (SUVAT)

Suggested reading Motion numerical Kinematics equations Numerical sets

Originally published at https://physicsteacher.in on December 20, 2020.

NIT Grad & IT prof. Writes and teaches HS physics. A motivational speaker and an active blogger with multiple niches. With lots of stories to tell.

NIT Grad & IT prof. Writes and teaches HS physics. A motivational speaker and an active blogger with multiple niches. With lots of stories to tell.