Question

equation of motion in graphical method for class 11th ISC

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

c. Equation for position velocity relation

Again consider graph in figure 7. We know that distance travelled s by a body in time t is given by the area under line AB which is area of trapezium OABC. So we have

Since OA+CB=u+v and OC=t, we thus have
s=(u+v)t2
From velocity time relation
t=v−ua
putting this t in equation for s we get
s=(u+v)2(v−ua)
or we have
v2=u2+2as
which is equation for position velocity relation.