Question

Derive the kinematic equations of motion.

Answer 1

Derivation of First Equation of Motion:

We know that the rate of change of velocity is the definition of body acceleration.

Let us assume a body that has a mass “m” and initial velocity “u”. Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”. Now we know that:

Acceleration = (Final Velocity-Initial Velocity) / Time Taken

a = v-u /t or at = v-u

v = u + at

Derivation of Second Equation of Motion:

Let the distance be “s”.

Distance = Average velocity × Time. Also, Average velocity (u+v)/2

Distance (s) = (u+v)/2 × t

Also, from v = u + at

s = (u+u+at)/2 × t = (2u+at)/2 × t

s = (2ut+at²)/2 = 2ut/2 + at²/2

or s = ut +½ at²

Derivation of Third Equation of Motion:

We have, v = u + at. Hence, we can write t = (v-u)/a

Also, we know that, Distance = average velocity × Time

Therefore, for constant acceleration we can write: Average velocity = (final velocity + initial velocty)/2 = (v+u)/2

Hence, Distance (s) = [(v+u)/2] × [(v-u)/a]

or s = (v² – u²)/2a

or 2as = v² – u²

or v² = u² + 2as.