Question

write derivations of equations of motion (if possible do try giving answer from ncert) pls guys don't send useless answer and do help me as fast as possible

Answer 1

Answer:

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Explanation:

v = u + a×t.

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

v² = u² + 2×a×s.

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 velocity)/2 = (v+u)/2

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

s = u×t + ½ ×a×t²

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