Question

any body describe the second equation of motion graphically

Answer 1

Let us consider that the object has travelled a distance s in time t under uniform acceleration a. In Fig. 7, the distance travelled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB.
Thus, the distance s travelled by the object is given by
s = area OABC (which is a trapezium)
s= area of the rectangle OADC + area of the triangle ABD
So,
s=OA×OC+12)AD×BD)
Substituting OA=u, OC=AD=t and BD=at, we get
s=(u×t)+12×(t×at)
or,
s=ut+12at2
which is the equation of position time relation

Answer 2

Please go through the answer
Hope it helps