Question

derive the third equation of motion v2 - u2=2as graphically​

Answer 1

• Here

u = initial velocity

v = final velocity

a = acceleration

t = time

s = distance

» Acceleration = $\dfrac{Change\:in\:velocity}{time\:taken}$

=> CD = $\dfrac{BD}{OA}$

=> CD = $\dfrac{AD\:-\:AB}{OA}$

=> a = $\dfrac{v\:-\:u}{t}$

=> at = v - u

=> t = $\dfrac{v\:-\:u}{a}$ ________(eq 1)

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Distance covered by body = Area of trapezium AOCDA

=> s = $\dfrac{1}{2}$ (OC + AD) (OA)

=> s = $\dfrac{1}{2}$ (u + v) (t)

=> s = $\dfrac{1}{2}$ (u + v) $(\dfrac{v\:-\:u}{a})$

=> s = $\dfrac{1}{2a}$ (v² - u²)

=> s = $\dfrac{ {v}^{2} \: - \: {u}^{2} }{2a}$

=> 2as = v² - u²

=> v² - u² = 2as

___________________[ANSWER]

Answer 2

Answer:

I can't understand

Explanation:

please ask properly