# Explain the kinematic equation in the graph.

The velocity-time graph for the uniformly accelerated body can be used to derive the three equations of motions.

Consider this velocity-time graph of a body moving with uniform acceleration. The body has an initial velocity 'u' at point A and then its velocity changes at a uniform rate from A to B in time ‘t’.
In other words, there is a uniform acceleration from A to B, and after time 't' its final velocity becomes 'v' which is equal to BC in the graph. The time 't' is represented by OC.
To complete the graph, a perpendicular CB is drawn from point C and a perpendicular AD is drawn parallel to OC. BE is perpendicular from point B to OE.

Now, starting with the first equation of motion.

1. Derivation of the first equation of motion: $v=u+at$ by graphical method -

Given:
Initial velocity of the body (u) = OA
Final velocity of the body (v) = BC

From the graph $BC=BD+DC$
$\therefore v=BD+DC$            $[\because v=BC]$
$v=BD+OA$                $[\because DC=OA]$
$v=BD+u$                    $[\because OA=u]$
$v=BD+u$  ------------------------- (i)

We know that the slope/gradient of the velocity-time graph is equal to acceleration 'a'.
Acceleration, a = slope of line AB
$a=\frac{BD}{AD}$
$a=\frac{BD}{t}$                            $[\because AD=OC=t]$
$BD=at$

Now, putting this value of BD in equation (i), we get:
$v=at+u$
Rearranging this equation to get:
$v=u+at$

Hence, the first equation of motion is derived by the graphical representation.

2. Derivation of the second equation of motion: $s=ut+\frac{1}{2}\times a{t}^{2}$ by graphical method -
Let's assume that the distance travelled's'  by the body in time 't' can be calculated by calculating the area under the velocity-time graph.

The area under the graph is equal to the area of OABC.

Thus,
Distance travelled = area of figure OABC
$=Area\ of\ triangle\ OADC+Area\ of\ triangle\ ABD$
$=(OA\times OC)+\left(\frac{1}{2}\times AD\times BD\right)$ $\left(\because Area\ of\ rectangle=length\times breadth\ and\ Area\ of\ triangle=\frac{1}{2}\times base\times height\right)$
$=(u\times t)+\left(\frac{1}{2}\times t\times at\right)$ $=ut+\frac{1}{2}a{t}^{2}$
So, distance travelled $s=ut+\frac{1}{2}a{t}^{2}$
$s=ut+\frac{1}{2}a{t}^{2}$

Hence, the second equation of motion is derived by the graphical representation.

3. Derivation of the third equation of motion: ${v}^{2}={u}^{2}+2as$ by graphical method -
We know that the area under the velocity-time graph gives us the distance travelled by the body.
The distance travelled 's' by the body in time 't' is given by the area of figure OABC (which is a trapezium).

Therefore,
Distance travelled = area of trapezium OABC
$s=\frac{(sum\ of\ parallel\ sides)\times (height)}{2}$  $\left[\because Area\ of \ trapezium=\frac{1}{2}\times (sum\ of\ parallel\ sides)\times (height)\right]$
$s=\frac{(OA+BC)\times OC}{2}$

$Now,\ OA+CB=u+v,\ and\ OC=t$
Substituting these values in above relation, we get –
$s=\frac{(u+v)\times t}{2}$

Now, to get the required equation we have to eliminate 't' from the above equation. And, this can be done by substituting the value of 't' from the first equation of motion.
$v=u+at$
$v-u=at$ $t=\frac{(v-u)}{a}$
Substituting this value in above equation, we get -
$s=\frac{(v+u)(v-u)}{2a}$
$2as={v}^{2}-{u}^{2}$
${v}^{2}={u}^{2}+2as$

Hence, the third equation of motion is derived by the graphical representation.

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Updated on: 10-Oct-2022

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