Question

Derivation of equation of motion by graphical method.

Answer 1

Answer:

We know that, BD = BC + CD (From figure)

⇒ v = BC + AO

⇒ v = BC + u

The slope of the v-t graph is said to be the acceleration.

Hence, v = u + at.

The area under the curve is said to be the displacement.

Hence, Displacement = Area of triangle ABC + Area of rectangle ACDO.

⇒ x = 1/2 bh + lb

⇒ x = 1/2 (v - u) t + ut

⇒ x = 1/2 at*t + ut [BC = (v - u) = at]

Hence, x = ut = 1/2 at².

We know that, x = Area of trapezium = 1/2 (sum of parallels)*h

⇒ x = 1/2(v + u)*t

Now, we know : a + (v - u)/t

So, t = (v - u)/a

Using the result,

⇒ x = 1/2(v + u)(v - u)/a

⇒ x = v² - u²/2a

Hence, v² - u² = 2ax.