Question

prove by graphical method v square = u square + 2as

Answer 1

Graphically Derivation of Third Equation of Motion :

Consider the velocity time graph of a body shown in attachment. The body has an initial Velocity u at point A and then velocity changes from A to B at a uniform rate in time t. There's uniform acceleration a from A to B,and after time t the final Velocity is v which is equal to BC. The time t is represented by OC. Through construction we drew perpendicular CB from point C,and AD parallel to OC. BE is perpendicular to the point B to OE.

Here,Distance Travelled s by a body in time t is given by the area of OABC which is a Trapezium.

★ Distance Travelled,s = Area of Trapezium OABC

→ s = (sum of parallel sides) × height/2

→ s = (OA + CB) × OC/2

Now,

OA + CB = u + v and OC = t

Put the values :

→ s = (u+v)× t/2....i)

Eliminate t from the above equation by obtaining value of t from first equation of motion.

★ v = u + at (First Equation of Motion)

→ at = v - u

→ t = (v-u)/a

Put the value of t in equation i)

→ s = (u+v)× t/2

→ s = (u+v) × (u-v)/2×a

→ s = (v² - u²)/2a

→ 2as = v² - u²

→ v² = u² + 2as

$\therefore$ Third Equation of Motion is derived by the Graphical Method.

Answer 2

We shall graphically prove the 3rd equation of motion : v² = u² + 2as

First , we shall draw the velocity versus time graph of an object travelling with constant acceleration. Let's assume that the ratio of velocity of the object is u and final velocity of the object is v.

We know that distance travelled by a object will be equal to the area enclosed by the velocity versus time graph.

Proof :

$\therefore\:$ Distance = Area of trapezium

=> s = ½ × (sum of parallel sides)×(dist.)

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

Representing "t" in form of velocity and acceleration :

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

$= > 2as = (v + u)(v - u)$

$= > 2as = {v}^{2} - {u}^{2}$

$= > {v}^{2} = {u}^{2} + 2as$

[ Hence Proved ]