Question

Derive the expression $S=ut+\frac{1}{2}a{t}^2$ using v-t graph

Answer 1

The second equation of motion:

The second equation of motion is given as;

$s=ut+\frac{1}{2}a{t}^2$

Where $\text{s}$ is the distance covered by the body, $\text{u}$ is the initial velocity, $v$ is the final velocity, $\text{t}$ is time and $\text{a}$ is the acceleration in the body

Graphical proof:

1. Let us Consider the linear motion of a body with an initial velocity($u$).
2. Let the body accelerate uniformly and acquire a final velocity($v$) after time $\text{t}$.

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 travelled in time t = area of the trapezium OABD

s=(12)×(OA+DB)×ODs = (\dfrac{1}{2}) \times (OA + DB) \times ODs=(21​)×(OA+DB)×OD

s=(12)×(u+v)×t s = (\dfrac{1}{2}) \times (u + v) \times t s=(21​)×(u+v)×t

Since $v=u+at$

s=(12)×(u+u+at)×ts = (\dfrac{1}{2}) \times (u + u + at) \times ts=(21​)×(u+u+at)×t

s=ut+(12)at2s = ut + (\dfrac{1}{2}) at^2s=ut+(21​)at2

This is the required expression.