Question

s = ut + 1/2.at²
please prove that this rule​

Answer 1

Answer:

Heya user............. ❤

Consider a velocity -time graph where the object is moving at constant acceleration.

( where you denotes initial velocity & final velocity)

Hence, the area of a velocity -time graph gives the displacement. That's why,

s = Area of AOCD + Area of ABD

= ut + 1/2 t (v-u)

= ut + 1/2 t *at [ v = u + at]

Hence, S = ut + 1/2 at^2

(proved)

Hope it’s helpful........ ☆(❁‿❁)☆

Answer 2

.

Let us consider that the object travelled a distance s in time t under uniform acceleration a. In the figure given I'm that attachment, the distance travelled by the object is traveled by the object is obtained by the area enclosed with OABC under the velocity-time graph AB.

Thus, the distance s travelled by the object is given by

s = area of OABC (which is a trapezium)

= area of the rectangle OADC + area of the triangle ABD

= OA × OC + ½(AD × BD)

Substituting OA = u, OC = AD = t and BD=at, we get

$s = u \times t + \frac{1}{2} (t \times at) \\ or \\ \: \red{s = ut + \frac{1}{2} a {t}^{2} }$