Question

Derivation of equation of motion ​

Answer 1

Derivation of the Equations of Motion

Frist equation

$\huge\purple{\mid{\fbox{\tt{v = u + at}}\mid}}$

Acceleration = Change in velocity/Time Taken

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

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

Therefore, we have:

v = u + at

second equation

$\huge\purple{\mid{\fbox{\tt{v² = u² + 2as}}\mid}}$

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

v² = u² + 2as

Third equation

$\huge\purple{\mid{\fbox{\tt{s = ut + ½at²}}\mid}}$

Let the distance be “s”.

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

Therefore, Distance (s) = (u+v)/2 × t

Also, from v = u + at, we have:

s = (u+u+at)/2 × t = (2u+at)/2 × t

s = (2ut+at²)/2 = 2ut/2 + at²/2

s = ut +½ at²

Answer 2

There are three equations of motion that can be used to derive components such as displacement(s), velocity (initial and final), time(t) and acceleration(a). ... First Equation of Motion : v=u+at. Second Equation of Motion : s=ut+\frac{1}{2}at^2. Third Equation of Motion : v^2=u^2+2as.