Derive the third equation of motion: $v^{2} - u^{2} = 2 a s$ .

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Solution

Step 1: To prove:
$v^{2} - u^{2} = 2 a s$
Where u is initial velocity, v is final velocity, a is the acceleration and s is the displacement, t is the time.
Step 2: Derivation:
For getting the third equation of motion, we use both of the equations of motion
$S = u t + \frac{1}{2} a t^{2}(1) (2 n d e q u a t i o n o f m o t i o n) v = u + a t___(2) (1 s t e q u a t i o n o f m o t i o n) t = (v - u) / a$
Substitute the value of t in equation (1), $S = u t + 1 / 2 a t^{2}$ ,
We get,
$S = u (\frac{v - u}{a}) + \frac{1}{2} a {(\frac{v - u}{a})}^{2}$
$\Rightarrow 2 a S = 2 u (v - u) + {(v - u)}^{2} \Rightarrow 2 a S = 2 u v - 2 u^{2} + v^{2} - 2 u v + u^{2} \Rightarrow 2 a S = v^{2} - u^{2} \Rightarrow v^{2} - u^{2} = 2 a s$
Hence, $v^{2} - u^{2} = 2 a s$

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Similar questions

Derive following equations for a uniformly accelerated motion :

(i) v=u+at

(ii) $S = u t + \frac{1}{2} a t^{2}$

(iii) $v^{2} = u^{2} + 2 a S$

where the symbols have their usual meanings.

(v^{2} - u^{2} = 2 a s)

(v = u + a t

s = u t + \frac{1}{2} a t^{2})

v^{2} = u^{2} + 2 a s

v^{2} - u^{2} = 2 a s

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