Question

drive Graphical the equation for position time relation for an object travelling distance S in time T under uniform acceleration

Answer 1

There are two kinematic equations connecting velocity and time.

Consider a velocity-time graph of an object which moves with uniform acceleration as shown in Fig below.

Object has initial velocity u at point A, which increases to value v, point B, in time t. The velocity changes at a uniform rate =a.

We draw perpendicular lines BC and BE from point B on time and velocity axes respectively.

Initial velocity is represented by OA, the final velocity is represented by BC and the time interval t is represented by OC. 
BD = BC – CD, is the change in velocity during time t.



Draw AD parallel to OC.

In the figure
BC = BD + DC = BD + OA
Now BC = v and OA = u

Therefore we get

v = BD + u
⇒ BD = v−u ........(1)

We know that rate of change of velocity is acceleration a

∴a=change in velocitytime taken

⇒a=BDAD=BDOC

Since OC = t we get

a=BDt
⇒BD=at .............(2)

From (1) and (2) we get

v=u+at .......(3)

From the Fig. the distance traveled s by the object is the area of the trapezium OABC.

Now
Area of OABC= area of the rectangle OADC + area of the Δ ABD
∴s= OA × OC + 12 (AD × BD) .......(4)

Substituting OA = u, OC = AD = tand BD = at, in (4) we get
s=u×t+12(t×at)

⇒s=ut+12at2 .......(5)

=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=.=

For sake of completeness
Kinematic Equation for Position-Velocity relation is established as follows.

In the velocity-time graph shown in the Fig. above the distance s travelled by the object in time t while moving under uniform acceleration ais the area of the trapezium OABC. From geometry we know that

Area of the trapezium OABC=OA+BC2× OC

Substituting OA = u, BC = v and OC = t we get

s=u+v2×t .......(5)

The velocity-time relation (3) can be rewritten as

t=v−ua ...........(6)

Substituting value of t from (6) in (5) we get

s=v+u2×v−ua
⇒2as=v2–u2