Question

An object starts linear motion with a velocity ‘u’ and under uniform acceleration ‘a’ acquires
a velocity ‘v’ in time‘t’. Draw its velocity time graph. From this graph obtain the following
equations:
1) S = ut + ½ at2
2) v2 – u
2 = 2as 3)v=u+at

Answer 1

⭐⭐⭐Your answer is in the attachment above ⭐⭐⭐

♣DC represents the initial velocity.

♣BC represents the final velocity.

♣Slope of AB represents acceleration.

♣And the area represents the distance traveled.

♣ Time is represented on x axis

#BE BRAINLY

Answer 2

Answer:

use the graph from the explanation given by the before mate

Step-by-step explanation:

s=ut +1/2at2

By position time relation:

s=area of trapezium OABC

=area of the rectangle OADC+area of the triangle ABD

=OA*OC+1/2 [AD*BD]

Substituting OA=u,

                     OC=AD=t and

                      BD=at,

we get, s=u*t+1/2[t*at]

           or

s=ut+1/2 at2

2}v2-u2=2as

distance [s]=area of the trapezium OABC

=[OA+BC]OC/2

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

s=[u+v]t/2----------------1

from the velocity time relation,

we get, s=v-u/a--------2

Using the equations 1 and 2, we get

s=[u+v][v-u]/2a

        or

2as= v2-u2

3)v=u+at

Let us draw AD parallel to OC. From

the graph we observe that

BC=BD+DC=BD+OA

Substituting BC=v and OA=u

we get, v=BD+u

            or

BD=v-u---------------1

From the velocity time graph, the acceleration of the object is given by,

a=change in velocity / time taken=BD/AD=BD/OC

Substituting OC=t,we get

a=BD/t

   or

BD=at----------------2

using the equations 1 and 2,we get

v=u+at