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# With the help of graph, derive the relation between:

a. velocity time

b. position time

c. position velocity (derive three equation of motion)

**Velocity-Time **

A car having initial velocity u is being subjected to a uniform acceleration a for a time period of t, after the time t the final velocity of the car is v.

Now we need to find the relation between v,u,a,t graphically.

Initial velocity u at A = OA

Final velocity is V

Slope of velocity time graph is equal to acceleration a.

$a\ =\ \frac{v\ -\ u}{t\ -\ 0}$

$\Longrightarrow \ v\ =\ u\ +\ at$

Position -Time

Consider the linear motion of a body with an initial velocity u. Let the body accelerate uniformly and acquire a final velocity v after time t. The velocity–time graph is a straight line AB as shown below.

At t = 0, initial velocity = u = OA

At t = t, final velocity = v = OC

The distance S traveled in time t = area of the trapezium OABD

$s\ =\ \frac{1}{2} \ \times \ ( OA\ +\ DB) \ \times \ OD$

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

Since v = u + at,

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

$s\ =\ ut\ +\ \frac{1}{2} at^{2}$

**Velocity- Position**

From, velocity time relation

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

substituting the value of 't' in equation $s\ =\ \frac{1}{2} \ \times \ ( u\ +\ v) \ \times \ t$

we get,

$s\ =\ \left(\frac{v\ +\ u}{2}\right) \ \times \ \left(\frac{v\ -\ u}{2}\right)$

=> v²=u²+2as