Question

from the velocity time graph derive an expression for position velocity relation​

Answer 1

The shaded area in above fig.

Let the initial velocity of the object = u

Let the object is moving with uniform acceleration, a.

Let object reaches at point B after time, t and its final velocity becomes, v

Draw a line parallel to x-axis DA from point, D from where object starts moving.

Draw another line BA from point B parallel to y-axis which meets at E at y-axis.

The distance covered by the object moving with uniform acceleration is given by the area of trapezium ABDO

Therefore,

 Area of trapezium ABDOE =1/2× (sum of parallel sides + distance between parallel sides)

Distance(S)=1 / 2 × (DO+BE)×OE

S= 1 / 2 × (u+v)×t...............(i)

we know that,

$a = \frac{v - u}{t}$

 

from above equation we can say,

$t = \frac{v - u}{a}$

.....(ii)

After substituting the value of t from equation (ii) in equation (i)

$s = \frac{(v + u)(v - u)}{2a}$

2aS=(u+v)(v−u)

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

Hence Proved.

Mark as brainliest

Appreciate me taking time to ans and follow me