Question

derive 2 and 3 equation of motion​

Answer 1

v² = u² + 2as

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

s = ut + ½at²

Let the distance be “s”. We know that

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

or s = ut +½ at²