Problem : Two cars, initially at a separation of 12 m, start simultaneously. First car “A”, starting from rest, moves with an acceleration 2
m
/
s
2
m /
s
2
, whereas the car “B”, which is ahead, moves with a constant velocity 1 m/s along the same direction. Find the time when car “A” overtakes car “B”.
Solution : We shall attempt this question, using concept of relative velocity. First we need to understand what does the event “overtaking” mean? Simply said, it is the event when both cars are at the same position at a particular time instant. Subsequently, the car coming from behind moves ahead.
Now relative velocity has an appropriate interpretation suiting to this situation. The relative velocity of “A” with respect to “B” means that the car “B” is stationary, while car “A” moves with relative velocity. Thus, problem reduces to a simple motion of a single body moving with a certain velocity (relative velocity) and covering a certain linear distance (here sepeartion of 12 m) with certain acceleration. Note that car “B” has no acceleration.
Relative velocity of “A” with respect to “B”, i.e.
v
AB
v
AB
, at the start of motion is :
v
AB
=
v
A
-
v
B
=
0
-
1
=
1
m
/
s
v
AB
=
v
A
-
v
B
=
0
-
1
=
1
m
/
s
Relative acceleration of “A” with respect to “B”, i.e. aAB is :
a
AB
=
a
A
-
a
B
=
2
-
0
=
2
m
/
s
2
a
AB
=
a
A
-
a
B
=
2
-
0
=
2
m
/
s
2
Applying equation of motion for a single body (remember the second body is rendered stationary),
x
=
u
t
+
1
2
a
t
2
⇒
12
=
-
1
x
t
+
1
2
x
2
x
t
2
⇒
t
2
-
t
-
12
=
0
⇒
t
2
+
3
t
-
4
t
-
12
=
0
⇒
(
t
+
3
)
(
t
-
4
)
=
0
x
=
u
t
+
1
2
a
t
2
⇒
12
=
-
1
x
t
+
1
2
x
2
x
t
2
⇒
t
2
-
t
-
12
=
0
⇒
t
2
+
3
t
-
4
t
-
12
=
0
⇒
(
t
+
3
)
(
t
-
4
)
=
0
Neglecting negative value,
⇒
t
=
4
s
⇒
t
=
4
s
Note : We can solve this problem without using the concept of relative velocity as well. The car “A” moves 12 m more than the distance covered by “B” at the time of overtake. If car “B” moves by “x” meters, then car “A” travels “x+12” meters. It means that :
Displacement of “A”
=
Displacement of “B”
+
12
Displacement of “A”
=
Displacement of “B”
+
12
1
2
a
t
2
=
v
t
+
12
1
2
a
t
2
=
v
t
+
12
Putting values,
⇒
1
2
x
2
x
t
2
=
1
x
t
+
12
⇒
t
2
-
t
-
12
=
0
⇒
1
2
x
2
x
t
2
=
1
x
t
+
12
⇒
t
2
-
t
-
12
=
0
This is the same equation as obtained earlier. Hence,
⇒
t
=
4
s
⇒
t
=
4
s
Evidently, it appears to be much easier when we do not use the concept of relative velocity. This is the case with most situations in one dimensional motion of two bodies. It is easier not to use concept of relative velocity in one dimensional motion to the extent possible. We shall learn subsequently, however, that the concept of relative velocity has an edge in analyzing two dimensional motion, involving two bodies.
