Question

Draw a velocity time graph for a body that has initial velocity u and is moving with acceleration a. Use it to derive v=u+at, s=ut+1/2at square, vsquare=usquare+2as

Answer 1

Consider the velocity-time graph of an object that moves under uniform acceleration as shown below in the figure 7.
From this graph, you can see that initial velocity of the object is u (at point A) and then it increases to v (at point B) in time t. The velocity changes at a uniform rate a.
Again from figure it is clear that time t is represented by OC , initial velocity u by OA and final velocity of object after time t by BC.

From graph as given in figure 7 it is clear that BC=BD+DC=BD+OA.
So we have
v=BD+u (1)
We should now find out the value of BD. From the velocity-time graph (Fig. 7), the acceleration of the object is given by

which gives, BD=at
putting this value of BD in equation 1 we get
v=u+at
which is the equation for velocity time relation.
b. Equation for position time relation

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.