Question

Derive the three equations of motion.

Answer 1

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).

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:

v = u + at

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

$s=(\frac{1}{2}AB\times BD)+(OA\times OC)$

Since BD = EA, the above equation becomes

$s=(\frac{1}{2}AB\times EA)+(u\times t)$

As EA = at, the equation becomes

$s=\frac{1}{2}\times at\times t+(u\times t)$

On further simplification, the equation becomes

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

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,

$s=\frac{1}{2}\left(Sumofparallelsides\right)\times \left(Height\right)$

$s=\frac{1}{2}(OA+CB)\times \left(OC\right)$

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

The above equation becomes

$s=\frac{1}{2}(u+v)\times t$

Now, since t = (v – u)/a

The above equation can be written as:

$s=\frac{1}{2}(u+v)\times \frac{(v-u)}{a}$

Rearranging the equation, we get

$s=\frac{v^2-u^2}{2a}$