Drive the Equation of motion by graphical motion.
Answers
Consider the velocity-time graph of an object
that moves under uniform acceleration as own in Fig. 8.8 (similar to Fig. 8.6, but now
with u ≠ 0). 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.
In Fig. 8.8, the perpendicular lines BC and
BE are drawn from point B on the time and
the velocity axes respectively, so that the
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, represents the change in velocity in time
interval t.
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 (8.8)
From the velocity-time graph (Fig. 8.8),
the acceleration of the object is given by
a =
Change in velocity
time taken
= BD BD = AD OC
Substituting OC = t, we get
a = BD
t
or BD = at (8.9)
Using Eqs. (8.8) and (8.9) we get
v = u + at
Let us consider that the object has travelled
a distance s in time t under uniform
acceleration a. In Fig. 8.8, 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)
= area of the rectangle OADC + area of
the triangle ABD
= OA × OC +
1
2 (AD × BD) (8.10)
Substituting OA = u, OC = AD = t and BD
= at, we get
s = u × t +
1 ( ) 2
t ×at
or s = u t +
1
2 a t 2
8.5.3 EQUATION FOR POSITION–VELOCITY
RELATION
From the velocity-time graph shown in
Fig. 8.8, the distance s travelled by the object
in time t, moving under uniform acceleration
a is given by the area enclosed within the
trapezium OABC under the graph. That is,
s = area of the trapezium OABC
= OA + BC ×OC
2
Substituting OA = u, BC = v and OC = t,
we get
2
u+v t
s (8.11)
From the velocity-time relation (Eq. 8.6),
we get
v–u
t =
a (8.12)
Using Eqs. (8.11) and (8.12) we have
[v+u] {v-u] s = 2a
or 2 a s = v2 – u2
