Derive the equations.Vector method of three equation of uniform accelerated motion

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Solution

In case of uniform accelerated motion in any direction:
Acceleration,
$\to a = \frac{Δ \to V}{Δ t} = \frac{B C}{A C} = \frac{\to V - {\to V}_{0}}{t} \to V = {\to V}_{0} + \to a t$ ..... (First equation of motion)
Now ,area of v-t graph =total displacement
$\to s = O A \times O D + \frac{1}{2} \times A C \times B C$
$= {\to V}_{o} t + \frac{1}{2} t (\to V - {\to V}_{0})$
$\to s {= \to V}_{0} t + \frac{1}{2} t \times \to a t$ ...(Second equation of motion)
Now $\to s$ = area of trapezium OABD
$= \frac{1}{2} \times (O A + B D) \times A C = \frac{1}{2} (\to V - {\to V}_{0}) t = \frac{1}{2} (\to V - {\to V}_{0}) ⎛ ⎜ ⎝ \frac{\to V - {\to V}_{0}}{\to a} ⎞ ⎟ ⎠ \to s = \frac{V^{2} - {V_{0}}^{2}}{2 \to a}$
$V^{2} - {V_{0}}^{2} = 2 \to a . \to s$ ...(Third equation of motion)

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