Derive the following equations for a uniformly accelerated motion —

(i) v = u + at

(ii) S = ut + $\dfrac{1}{2}$ at²

(iii) v² = u² + 2aS

where the symbols have their usual meanings.

(i) By definition,

$\text {Acceleration} = \dfrac {\text {change in velocity}}{\text {time taken}} \\[0.5em] \Rightarrow \text {Acceleration} = \dfrac {\text {final velocity - initial velocity}}{\text {time taken}} \\[0.5em]$

Hence,

$\text {a} = \dfrac{\text{v} - \text{u}}{\text {t}} \\[0.5em] \text {at} = \text{v} - \text{u} \\[0.5em] \Rightarrow \text {v} = \text{u} + \text{at} \\[0.5em]$

Hence, we get,

v = u + at     [Eq. 1]

(ii) By definition,

$\text {Distance travelled} = \text {Avg. Velocity} \times \text{time} \\[1em] \text {Distance travelled} = \Big(\dfrac{\text {Initial Vel.} + \text {Final Vel.}}{2}\Big) \times \text{time}$

or

$\text{S} = \Big(\dfrac {\text{u} + \text{v}}{2}\Big) \times \text{t}$

But, from Eq. 1, v = u + at

Therefore,

$\text{S} = \Big(\dfrac{\text {u + ( u +at)}}{2}\Big) \times \text {t} \\[0.5em] \text{S} = \Big(\dfrac{\text {2u + at}}{2}\Big) \times \text {t} \\[0.5em] \Rightarrow \text{S} = \text {ut} + \dfrac{1}{2} \text {at}^2 \\[0.5em]$

Hence, we get,
S = ut + $\bold{\dfrac{1}{2}}$ at²

(iii) By definition,

$\text {Distance travelled} = \text {Avg. Velocity} \times \text{time} \\[0.5em] \Rightarrow \text{S} = \Big(\dfrac {\text{u} + \text{v}}{2}\Big) \times \text{t}$

But, from Eq. 1, v = u + at

or t = $\dfrac{\text{v}-\text{u}}{\text{a}}$

Therefore,

$\text{S} = \Big(\dfrac {\text{v} + \text{u}}{2}\Big) \times \Big(\dfrac{\text{v}-\text{u}}{a}\Big) \\[1em] \Rightarrow \text{S} = \dfrac{\text{v}^2 - \text{u}^2}{2\text{a}}$

So, v² - u² = 2aS

Hence,

v² = u² + 2aS