Derivation Of Equation Of Motion

Derivation of First Equation of Motion by Graphical Method

The first equation of motion can be derived using a velocity-time graph for a moving object with an initial velocity of u, final velocity v, and acceleration a.

In the above graph,

• The velocity of the body changes from A to B in time t at a uniform rate.
• BC is the final velocity and OC is the total time t.
• A perpendicular is drawn from B to OC, a parallel line is drawn from A to D, and another perpendicular is drawn from B to OE (represented by dotted lines).

The following details are obtained from the graph above:

The initial velocity of the body, u = OA

The final velocity of the body, v = BC

From the graph, we know that

BC = BD + DC

Therefore, v = BD + DC

v = BD + OA (since DC = OA)

Finally,

v = BD + u (since OA = u) (Equation 1)

Now, since the slope of a velocity-time graph is equal to acceleration a,

So,

a = slope of line AB

a = BD/AD

Since AD = AC = t, the above equation becomes:

BD = at (Equation 2)

Now, combining Equation 1 & 2, the following is obtained:

Derivation of First Equation of Motion by Calculus Method

Since acceleration is the rate of change of velocity, it can be mathematically written as:

\(\begin{array}{l}a=\frac{dv}{dt}\end{array} \)

Rearranging the above equation, we get

\(\begin{array}{l}adt=dv\end{array} \)

Integrating both the sides, we get

\(\begin{array}{l}\int_{0}^{t}adt=\int_{u}^{v}dv\end{array} \)

\(\begin{array}{l}at=v-u\end{array} \)

Rearranging, we get

\(\begin{array}{l}v=u+at\end{array} \)

Derivation of Second Equation of Motion

For the derivation of the second equation of motion, consider the same variables that were used for derivation of the first equation of motion.

 

Derivation of Second Equation of Motion by Algebraic Method

Velocity is defined as the rate of change of displacement. This is mathematically represented as:

\(\begin{array}{l}Velocity=\frac{Displacement}{Time}\end{array} \)

Rearranging, we get

\(\begin{array}{l}{Displacement}=Velcoity\times Time\end{array} \)

If the velocity is not constant then in the above equation we can use average velocity in the place of velocity and rewrite the equation as follows:

\(\begin{array}{l}{Displacement}=(\frac{Initial\,Velocity + Final\,Velocity}{2})\times Time\end{array} \)

Substituting the above equations with the notations used in the derivation of the first equation of motion, we get

\(\begin{array}{l}s=\frac{u+v}{2}\times t\end{array} \)

From the first equation of motion, we know that v = u + at. Putting this value of v in the above equation, we get

\(\begin{array}{l}s=\frac{u+(u+at))}{2}\times t\end{array} \)

\(\begin{array}{l}s=\frac{2u+at}{2}\times t\end{array} \)

\(\begin{array}{l}s=(\frac{2u}{2}+\frac{at}{2})\times t\end{array} \)

\(\begin{array}{l}s=(u+\frac{1}{2}at)\times t\end{array} \)

On further simplification, the equation becomes:

\(\begin{array}{l}s=ut+\frac{1}{2}at^2\end{array} \)

Derivation of Second Equation of Motion by Graphical Method

From the graph above, we can say that

Distance travelled (s) = Area of figure OABC = Area of rectangle OADC + Area of triangle ABD

\(\begin{array}{l}s=(\frac{1}{2}AB\times BD)+(OA\times OC)\end{array} \)

Since BD = EA, the above equation becomes

\(\begin{array}{l}s=(\frac{1}{2}AB\times EA)+(u\times t)\end{array} \)

As EA = at, the equation becomes

\(\begin{array}{l}s=\frac{1}{2}\times at\times t+ut\end{array} \)

On further simplification, the equation becomes

\(\begin{array}{l}s=ut+\frac{1}{2}at^2\end{array} \)

Derivation of Second Equation of Motion by Calculus Method

Velocity is the rate of change of displacement.

Mathematically, this is expressed as

\(\begin{array}{l}v=\frac{ds}{dt}\end{array} \)

Rearranging the equation, we get

\(\begin{array}{l}ds=vdt\end{array} \)

Substituting the first equation of motion in the above equation, we get

\(\begin{array}{l}ds=(u+at)dt\end{array} \)

\(\begin{array}{l}ds=(u+at)dt=(udt+atdt)\end{array} \)

On further simplification, the equation becomes:

\(\begin{array}{l}\int_{0}^{s}ds=\int_{0}^{t}udt+\int_{0}^{t}atdt\end{array} \)

\(\begin{array}{l}=ut+\frac{1}{2}at^2\end{array} \)

Derivation of Third Equation of Motion

For the derivation of the third equation of motion, consider the same variables that were used for the derivation of the first and second equations of motion.

Derivation of Third Equation of Motion by Algebraic Method

We know that displacement is the rate of change of position of an object. Mathematically, this can be represented as:

\(\begin{array}{l}Displacement=(\frac{Initial\,Velocity+Final\,Velocity}{2})\times t\end{array} \)

Substituting the standard notations, the above equation becomes

\(\begin{array}{l}s=(\frac{u+v}{2})\times t\end{array} \)

From the first equation of motion, we know that

\(\begin{array}{l}v=u+at\end{array} \)

Rearranging the above formula, we get

\(\begin{array}{l}t=\frac{v-u}{a}\end{array} \)

Substituting the value of t in the displacement formula, we get

\(\begin{array}{l}s=(\frac{v+u}{2})(\frac{v-u}{a})\end{array} \)

\(\begin{array}{l}s=(\frac{v^2-u^2}{2a})\end{array} \)

\(\begin{array}{l}2as={v^2-u^2}\end{array} \)

Rearranging, we get

\(\begin{array}{l}v^2=u^2+2as\end{array} \)

Derivation of Third Equation of Motion by Graphical Method

From the graph, we can say that

The total distance travelled, s is given by the Area of trapezium OABC.

Hence,

S = ½ (Sum of Parallel Sides) × Height

S=(OA+CB)×OC

Since, OA = u, CB = v, and OC = t

The above equation becomes

S= ½ (u+v) × t

Now, since t = (v – u)/ a

The above equation can be written as:

S= ½ ((u+v) × (v-u))/a

Rearranging the equation, we get

S= ½ (v+u) × (v-u)/a

S = (v²-u²)/2a

Third equation of motion is obtained by solving the above equation:

v2 = u2+2aS

Derivation of Third Equation of Motion by Calculus Method

We know that acceleration is the rate of change of velocity and can be represented as:

\(\begin{array}{l}a=\frac{dv}{dt}\end{array} \)

(1)

We also know that velocity is the rate of change of displacement and can be represented as:

\(\begin{array}{l}v=\frac{ds}{dt}\end{array} \)

(2)

Cross multiplying (1) and (2), we get

\(\begin{array}{l}a\frac{ds}{dt}=v\frac{dv}{dt}\end{array} \)

\(\begin{array}{l}\int_{0}^{s}ads=\int_{u}^{v}vdv\end{array} \)

\(\begin{array}{l}as=\frac{v^2-u^2}{2}\end{array} \)

\(\begin{array}{l}v^2=u^2+2as\end{array} \)

This is how we derive the three equations of motion by algebraic, graphical and calculus method.
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