Question

Prove that S=ut + ½ at2​

Answer 1

Let us Consider the linear motion of a body with an initial velocity u. Let the body accelerate uniformly and acquire a final velocity v after time t. Here, The velocity-time graph is a straight line AB

So, According to the velocity-time graph is

At t=0,

initial velocity =u=OA

At t=t,

final velocity =v=OC

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

s=( ½)×(OA+DB)×OD

s=( ½)×(u+v)×t

Since v=u+at

s=( ½)×(u+u+at)×t

S = ut + (½) at²

This is the required expression

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Answer 2

Explanation:

By using graphical method

Distance = Area under the slope

Area under the slope = area of Triangle + area of rectangle

area of Triangle = ½ * base * height

therefore ,

1/2 * Ad * (Bd - cd)

½ * t * ( v- u )

similarly ,

Area of rectangle = L * B

Now ,

oc * ao

t * u

Now Add the values .......

s = ½ * t * ( v-u ) + t * u

s = ½ * t * ( at) + ut • we can take v - u is at by the formula of accleration = (v - u)/t = v- u = at

s = ut + ½ at²

Hence , proved .....