Question
Derive the equations of motion by graphical method.
✓
Answer
An object is in motion with initial velocity u attains a final velocity v in time t due to acceleration a, with displacement S.
Let us try to derive these equations by graphical method. Equations of motion from velocity – time graph:
Graph shows the change in velocity with time for an uniformly accelerated object. The object starts from the point $D$ in the graph with velocity $u$. Its velocity keeps increasing and after time $t$ it reaches the point B on the graph.
The initial velocity of the object $= u = OD = EA$
The final velocity of the object $=v=O C=E B$
Time $= t = OE = DA$
Also from the graph we know that, $A B=D C$
For First equation of motion
By definition, acceleration $=\frac{\text { change in velocity }}{\text { time }}$
$=\frac{(\text { final velocity }- \text { initial velocity })}{\text { time }}$
$=\frac{(\text { final velocity }- \text { initial velocity })}{\text { time }}$
$=\frac{(O C-O D)}{O E}=\frac{D C}{O E}$
$a=\frac{D C}{t}$
$DC = AB =$ at
From the graph $E B=E A+A B$
$v=u+a t$
This is first equation of motion.
For Second equation of motion
From the graph the distance covered by the object during time $t$ is given by the area of quadrangle DOEB
$s$ = area of the quadrangle DOEB
$s=$ area of the rectangle DOEA + area of the triangle DAB
$=(A E \times O E)+(1 / 2 \times A B \times D A)$
$s=u t+1 / 2 a t^2 \ldots(2)$
This is second equation of motion.
For Third equation of motion
From the graph the distance covered by the object during time $t$ is given by the area of the quadrangle DOEB. Here DOEB is a trapezium. Then
$
\begin{array}{l}
s=\text { area of trapezium DOEB } \\
=1 / 2 \times \text { sum of length of pa } \\
=1 / 2 \times(O D+B E) \times O E \\
s=1 / 2 \times(u+v) \times t
\end{array}
$
$
=1 / 2 \times \text { sum of length of parallel side } \times \text { distance between parallel sides }
$\
Since $a=$ Since $a=(v-u) /$ tor $t=(v-u) / a$
Therefore $=1 / 2 \times(v+u) \times(v-u) / a$.
$
\text { 2as }=v^2=u^2+2 a s
$
$
v^2=u^2+2 \text { as .........(3) }
$
This is third equation of motion.
Let us try to derive these equations by graphical method. Equations of motion from velocity – time graph:
Graph shows the change in velocity with time for an uniformly accelerated object. The object starts from the point $D$ in the graph with velocity $u$. Its velocity keeps increasing and after time $t$ it reaches the point B on the graph.
The initial velocity of the object $= u = OD = EA$
The final velocity of the object $=v=O C=E B$
Time $= t = OE = DA$
Also from the graph we know that, $A B=D C$
For First equation of motion
By definition, acceleration $=\frac{\text { change in velocity }}{\text { time }}$
$=\frac{(\text { final velocity }- \text { initial velocity })}{\text { time }}$
$=\frac{(\text { final velocity }- \text { initial velocity })}{\text { time }}$
$=\frac{(O C-O D)}{O E}=\frac{D C}{O E}$
$a=\frac{D C}{t}$
$DC = AB =$ at
From the graph $E B=E A+A B$
$v=u+a t$
This is first equation of motion.
For Second equation of motion
From the graph the distance covered by the object during time $t$ is given by the area of quadrangle DOEB
$s$ = area of the quadrangle DOEB
$s=$ area of the rectangle DOEA + area of the triangle DAB
$=(A E \times O E)+(1 / 2 \times A B \times D A)$
$s=u t+1 / 2 a t^2 \ldots(2)$
This is second equation of motion.
For Third equation of motion
From the graph the distance covered by the object during time $t$ is given by the area of the quadrangle DOEB. Here DOEB is a trapezium. Then
$
\begin{array}{l}
s=\text { area of trapezium DOEB } \\
=1 / 2 \times \text { sum of length of pa } \\
=1 / 2 \times(O D+B E) \times O E \\
s=1 / 2 \times(u+v) \times t
\end{array}
$
$
=1 / 2 \times \text { sum of length of parallel side } \times \text { distance between parallel sides }
$\
Since $a=$ Since $a=(v-u) /$ tor $t=(v-u) / a$
Therefore $=1 / 2 \times(v+u) \times(v-u) / a$.
$
\text { 2as }=v^2=u^2+2 a s
$
$
v^2=u^2+2 \text { as .........(3) }
$
This is third equation of motion.
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