Question

Derive third equation of motion v-u^2=2aS graphically?​

Answer 1

Explanation:

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Question

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

Answer · 74 votes

• Hereu = initial velocityv = final velocitya = accelerationt = times = 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)____________________________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]

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Answer 2

Consider the velocity-time graph for a body having some non zero initial velocity at time t = 0.

image

u: velocity at time t1

v: velocity at time t2

a: uniform acceleration of the body along the straight line

Displacement covered during the time interval t2 - t1 = Area ABt1 t2

S = Area of triangle ABA’ + Area of rectangle AA’t2 t1 = Area of trapezium

S = ½ x Sum of parallel sides x Perpendicular distance

image

… (Equation 1)

From the first equation of motion, v = u + at;

image

Substituting in equation 1, we get

image

or, v2 – u2 = 2as

which is the required equation of motion.