Question

derive equation for position velocity relation v²-u²=2as​

Answer 1

Explanation:

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 (ABt₁t₂)

⇒ S = Area of triangle ABA’ + Area of rectangle AA’t₂t₁  = Area of trapezium

⇒ S = ½  x Sum of parallel sides x Perpendicular distance

⇒ $\frac{1}{2} \frac{(v+u)}{t}$

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

⇒  $t = \frac{v-u}{a}$

Substituting in equation 1, we get  

⇒ $\frac{1}{2a} (v-u)(v+u)$

⇒ $v^{2} - u^{2} = 2as$

which is the required equation of motion.

Hope this helps.....

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