Problem : The car “B” is ahead of car “A” by 10 km on a straight road at a given time. At this instant, car “B” turns left on a road, which makes right angle to the original direction. If both cars are moving at a speed of 40 km/hr, then find the closest approach between the cars and the time taken to reach closest approach.
Solution : Closest approach means that the linear distance between the cars is mimimum. There are two ways to handle this question. We can proceed in the conventional manner, considering individual velocity as observed from ground reference. We use calculus to find the closest approach. Alternatively, we can use the concept of relative velocity and determine the closest approach. Here, we shall first use the relative velocity technique. By solving in two ways, we shall reinforce the conceptual meaning of relative velocity and see that its interpretation in terms of a stationary reference is a valid conception.
The car “A” moves with certain relative velocity with respect to “B”, which is given by :
v
A
B
=
v
A
−
v
B
v
A
B
=
v
A
−
v
B
Since they are moving with same speed and at right angle to each other, relative velocity is as shown in the figure. Using pythagoras theorem, the magnitude of relative velocity is :
v
A
B
=
v
2
+
v
2
=
40
2
+
40
2
=
40
2
m
/
s
v
A
B
=
v
2
+
v
2
=
40
2
+
40
2
=
40
2
m
/
s
Its direction with respect to original direction is :
tan
θ
=
40
40
=
1
tan
θ
=
40
40
=
1
θ
=
45
0
θ
=
45
0
As we have discussed, relative velocity can be conceptually interpreted as if car “B” is stationary and car “A” is moving with a velocity
40
2
40
2
making an angle 45°. Since perpendicular drawn from the position of “B” (which is stationary) to the path of motion of “A” is the shortest linear distance and hence the closest approach. Therefore, closets approach by trigonometry of Δ ABC,
r
min
=
A
B
sin
45
0
=
10
x
1
2
=
5
2
k
m
r
min
=
A
B
sin
45
0
=
10
x
1
2
=
5
2
k
m
We should understand here that car “A” is moving in the direction as shown from the perspective of car “B”.
The time taken by car “A” to travel
5
2
5
2
km with a velocity of
40
2
40
2
m/s along the direction is :
t
=
5
2
40
2
=
1
8
h
r
=
7.5
min
t
=
5
2
40
2
=
1
8
h
r
=
7.5
min
Now, let us now attempt to analyze motion from the ground’s perpective. The figure here shows the positions of the car and linear distance between them at a displacement of 1 km. The linear distance first decreases and then increases. At a given time, the linear distance between the cars is :
r
=
{
10

x
2
+
y
2
}
r
=
{
10

x
2
+
y
2
}
For minimum distance, first time derivative is equal to zero. Hence,
⇒
d
r
d
t
=
−
2
10
−
x
d
x
d
t
+
2
y
d
y
d
t
2
{
10

x
2
+
y
2
}
=
0
⇒
d
r
d
t
=
−
2
10
−
x
d
x
d
t
+
2
y
d
y
d
t
2
{
10

x
2
+
y
2
}
=
0
⇒
−
2
10
−
x
d
x
d
t
+
2
y
d
y
d
t
=
0
⇒
−
2
10
−
x
d
x
d
t
+
2
y
d
y
d
t
=
0
⇒

10
+
x
+
y
=
0
⇒

10
+
x
+
y
=
0
⇒
x
+
y
=
10
⇒
x
+
y
=
10
Since cars are moving with same speed, x = y. Hence,
x
=
y
=
5
k
m
x
=
y
=
5
k
m
The closest approach, therefore, is :
r
min
=
{
10

x
2
+
y
2
}
=
{
10

5
2
+
5
2
}
=
5
2
r
min
=
{
10

x
2
+
y
2
}
=
{
10

5
2
+
5
2
}
=
5
2