Question

derive of s=ut+1/2at2​

Answer 1

Answer:

At t = 0, initial velocity = u = OA  

At t = t, final velocity = v = OC

The distance S travelled in time t = area of the trapezium OABD

s = (1/2) x (OA + DB) × OD

s = (1/2) x (u + v) × t  

Since v = u + at,

s = (1/2) x (u + u + at) × t

s = ut + (1/2) at2  

Hope it helps :)

Answer 2

Please click on the brainliest option below :)

To prove $\sf{S = ut + \dfrac{1}{2}a {t}^{2}}$ Where S is the distance.

Distance travelled = Area of trapezium ABCE

$\sf{S = \dfrac{1}{2} (AB + CE) \times AE}$

${ \sf{S = \dfrac{1}{2} (u + v) \times t}}$

${ \sf{S = \dfrac{1}{2} (u + u + at) \times t}}$

$\sf{ (\because v = u + at)}$

$\sf{S = \dfrac{1}{2} (2u + at) \times t}$

$\sf{ \therefore S = ut + \dfrac{1}{2} a {t}^{2} }$