Question

derive equation of motion i)v2-u2=2as
ii)v=u+at​

Answer 1

Answer:

First Equation: v = u + at

Final velocity = Initial velocity + Acceleration × Time

Graphical Derivation

Suppose a body has initial velocity ‘u’ (i.e., velocity at time t = 0 sec.) at point ‘A’ and this velocity changes to ‘v’ at point ‘B’ in ‘t’ secs. i.e., final velocity will be ‘v’.

For such a body there will be an acceleration.

a = Change in velocity/Change in Time

⇒ a = (OB - OA)/(OC-0) = (v-u)/(t-0)

⇒ a = (v-u)/t

⇒ v = u + at

Second Equation: s = ut + ½ at2

Distance travelled by object = Area of OABC (trapezium)

= Area of OADC (rectangle) + Area of ∆ABD

= OA × AD + ½ × AD × BD

= u × t + ½ × t × (v – u)

= ut + ½ × t × at

⇒ s = ut + ½ at2 (∵a = (v-u)/t)

Third Equation: v2 = u2 + 2as

s = Area of trapezium OABC

⇒ v2 = u2 + 2as

Explanation:

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Answer 2

Answer:

Explanation:hey this is correct ! Hope it helps