# Physics 9th: Motion | Derivation of Equations of Motion by Graphical Method

## Physics 9th: Motion | Derivation of Equations of Motion by Graphical Method

### TO DERIVE v = u + at BY GRAPHICAL METHOD

This is a graph of uniform acceleration with ‘u’ as initial velocity and ‘v’ as final velocity.

Initial velocity = u = OP

Final velocity = v = RN

= RQ + QN

v = u + QN    …..(i)

Acceleration, a = slope of line PN

$\displaystyle a=\frac{QN}{PQ}=\frac{QN}{t}$

$\displaystyle QN=at$        …..(ii)

Putting the value of QN from equation (ii) into equation (i), we get v = u + at

### TO DERIVE $\displaystyle s=ut+\frac{1}{2}a{{t}^{2}}$ BY GRAPHICAL METHOD

In the above speed-time graph, the distance travelled is given by

Distance travelled = Area of figure OPNR

= Area of DPNQ + Area of rectangle OPQR

(1)     Area of triangle PNQ = $\displaystyle \frac{1}{2}\times base\times height$

= $\displaystyle \frac{1}{2}\times PQ\times NQ$ $\displaystyle =\frac{1}{2}\times t\times (v-u)$

= $\displaystyle \frac{1}{2}\times t\times at$    [As v = u + at and v u = at]

Area of DPNQ = $\displaystyle \frac{1}{2}a{{t}^{2}}$

(2)    Area of rectangle OPQR = OP ´ PQ

= u ´ t

= ut

Distance travelled = Area of DPNQ + Area of rectangle OPQR

$\displaystyle S=\frac{1}{2}a{{t}^{2}}+ut$

$\displaystyle S=ut+\frac{1}{2}a{{t}^{2}}$

### TO DERIVE $\displaystyle {{v}^{2}}-{{u}^{2}}=2aS$ BY GRAPHICAL METHOD

In the above speed time graph distance travelled (S) = Area of trapezium OPNR

$\displaystyle S=\frac{1}{2}\times$ (sum of parallel sides) ´ height

$\displaystyle S=\frac{1}{2}\times$ (OP + RN) ´ OR

$\displaystyle S=\frac{1}{2}\times$ (u + v) ´ t

$\displaystyle S=\frac{1}{2}\times$ (v + u) t            …(i)

But v = u + at

at = v u

$\displaystyle t=\left( \frac{v-u}{a} \right)$                … (ii)

Putting this value of ‘t’ from equation (ii) into equation (i) we get that ___

$\displaystyle S=\frac{1}{2}\times$ (v + u) $\displaystyle \left( \frac{v-u}{a} \right)$

$\displaystyle S=\frac{\left( v+u \right)\left( v-u \right)}{2a}$

$\displaystyle S=\frac{{{v}^{2}}-{{u}^{2}}}{2a}$        [As (a + b) (a b) = $\displaystyle {{a}^{2}}-{{b}^{2}}$]

2as = $\displaystyle {{v}^{2}}-{{u}^{2}}$

$\displaystyle {{v}^{2}}-{{u}^{2}}=2as$

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