Derive the kinematic equations of motion.
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.