Equations of Motion || Physics || Class 9

Equation of motion describes the basic concepts of the motions of an object such as position, velocity or the acceleration of an object at various times whenever certain object is moving with an uniform acceleration. They are basically meant to define the physical system behaviour in terms of motion as a function of time.

Let us consider an object of which initial velocity is \(u \) and is moving with an uniform acceleration \(a \) for time \(t \) and travelled a distance \(s \) when \(v \) is the final velocity. Then a set of equations can be made describing their relations as follows-

\[\ v = u + at ~~(velocity-time~ relation)~ \] \[\ s = ut + \frac{1}{2}at^2 ~~(position-time~ relation)~ \] \[\ 2as = v^2-u^2 ~~(position-velocity~ relation)~ \]

We can derive them in numerous ways but here we'll see the derivation only by algebraic method and graphical method.

First Equation of Motion

\[\ v = u + at \] Derivation by Algebraic Method:
If an object is moves with an initial velocity \(u \) and final velocity \(v \) in time \(t \) then by definiotion of acceleration, we have \[\ a = \frac{v-u}{t} \] \[\ ⇨ at = v-u \] \[\ ⇨ v = u+at \] which is the required First Equation of motion.

Derivation by Graphical Method

Let us consider an object is moves from point A where the initial velocity of the object is \(u \) to point B it increases to \(v \) which is the final velocity in time interval \(t \) at uniform rate, so the velocity changes from A to B at an uniform rate \(a \) (as shown in fig.1). Let us draw the perpendicular BC and BE from the two axes of the graph of which O is the origin. Also, we draw AD parallel to OC. So, from the graph, the following details can be obtained,

initial velocity \(= u = \) OA
and final velocity \(= v = \) BC
Also, we know, BC = BD+DC
⸫ \(v = \) BD+DC = BD+OA [⸪ OA=DC]
Finally, \(v = \) BD+ \(u \) [⸪ OA= \(u \)] ...(1)
⇨ BD = \(v - u \)
Now, the slope of the velocity-time graph is equal to the acceleration.
So, \(a = \) Slope of the line AB
\( = \frac{BD}{AD} \) ...(2)
Since, AD = AC = \(t \)
⸫ (1)⇨ \( at= \) BD ...(3)
Combining (1)&(2), we get,
\(v=u+at \)
Which is the required First Equation of Motion.

Second Equation of Motion

\[\ s = ut + \frac{1}{2}at^2 \] Derivation by Algebraic Method:
We know that velocity can mathematically be defined as \[\ velocity = \frac{displacement}{time} \] \[\ ⇨ displacement = velocity × time \] But, if the velocity is not constant, then average velocity can be used instead of velocity, i.e., \( displacement = (\frac{initial~ velocity+final~ velocity}{2})×time \) So, using standard notation, we get
\[\ s = (\frac{u+v}{2})×t \]
\[\ ⇨ s = (\frac{u+(u+at)}{2})×t \]

[⸪ \(v=u+at \), First Equation of Motion]

\[\ ⇨ s = (\frac{2u+at}{2})×t \]
\[\ ⇨ s = (\frac{2u}{2}+\frac{at}{2})×t \]
\[\ ⇨ s = (u+\frac{at}{2})×t \]
\[\ ⇨ s = ut+\frac{1}{2}at^2 \]
Which is the required second equation of motion.

Derivation by Graphical Method

Let us consider an object has travelled a distance \(s \) in time \(t \) with an uniform acceleration \(a \) as shown in the figure. So, the distance travelled by the object obtained by the area enclosed within the trapezium OABD (as shown in fig. 2).
⸫ \(s \) = Area of the trapezium OABD
\( ⇨ s \) = Area of the rectangle OADC + Area of the traingle ABC
\( ⇨ s \) = (AO×OD) + \(\frac{1}{2}\)(AX×BC) ...(1)
So, from the graph, the following details can be obtained,

AO = \(u \), OD = AC = \(t \)
BC = BD \(- \) CD = \(v -u \)

⸫ (1)⇨ \(s = \frac{1}{2}[t×(v-u)]+ut \)
\(⇨ s = \frac{1}{2}t[v-u]+ut \)
\(⇨ s = \frac{1}{2}t[at]+ut \) [ \(at = v-u \)]
\(⇨ s = \frac{1}{2}at^2+ut \)
\(⇨ s = ut+\frac{1}{2}at^2 \)
Which is the required second equation of motion.

Third Equation of Motion

\[\ 2as = v^2-u^2 \] Derivation by Algebraic Method:
We have, \( displacement = (\frac{initial~ velocity+final~ velocity}{2})×time \) \[\ ⇨ s = (\frac{u+v}{2})×t~ ...(1) \] Also, we have from the Fisrt equation of motion, \[\ v = u+at ⇨ t = \frac{v-u}{a} \] \[\ ⸫~ (1) ⇨ s = (\frac{u+v}{2})(\frac{v-u}{a}) \] \[\ ⇨ s = \frac{v^2-u^2}{2a} \] \[\ ⇨ 2as = v^2 - u^2 \] Which is the required third equation of motion.

Derivation by Graphical Method

Let us consider an object is travelled a distance \(s \) with uniform acceleration \(a \) in time \(t \) then this distance can be represented by area enclosed within the trapezium OABC (as shown in fig.3).
So, the total distance travelled,
\(s \) = Area of the trapezium
\(s = \frac{1}{2} \) ×(Sum of the parallel sides)×(distance between them)
\(s = \frac{1}{2} \) ×(OA+CB)× OC ...(1)
So, from the graph, the following details can be obtained,

\(u \) = OA, \(v \) = BC and \(t \) = OC

⸫ (1)⇨ \(s = \frac{1}{2}×(u+v)×t \)
\(⇨ s = \frac{1}{2}×(u+v)×(\frac{v-u}{a}) \) [⸪ \(t = \frac{v-u}{a} \)]
\(⇨ s = \frac{1}{2}(v+u)(\frac{v-u}{a}) \)
\(⇨ s = \frac{1}{2a}(v^2-u^2) \) [⸪ \((v+u)(v-u)=v^2-u^2 \)]
\(⇨ v^2 = u^2 +2as \)
Which is the required Third Equation of Motion.