Solution : We draw the figure to illustrate the situation. Here, car travels the linear distance (AB + CD) along the direction in which it moves, by which time the pedestrian travels the linear distance BD. Let pedestrian travels at a speed “v” along BD, which makes an angle “θ” with the direction of car.
We must understand here that there may be number of combination of angle and speed for which pedestrian will be able to safely cross before car reaches. However, we are required to find the minimum speed. This speed should, then, correspond to a particular value of θ.
We can also observe that pedestrian should move obliquely. In doing so he/she gains extra time to cross the road.
From triangle BCD,
tan
(
90

θ
)
=
cot
θ
=
CD
BC
=
CD
2
⇒
CD
=
2
cot
θ
tan
(
90

θ
)
=
cot
θ
=
CD
BC
=
CD
2
⇒
CD
=
2
cot
θ
Also,
cos
(
90

θ
)
=
sin
θ
=
BC
BD
=
2
BD
⇒
BD
=
2
sin
θ
cos
(
90

θ
)
=
sin
θ
=
BC
BD
=
2
BD
⇒
BD
=
2
sin
θ
According to the condition given in the question, the time taken by car and pedestrian should be equal for the situation outlined above :
t
=
4
+
2
cot
θ
8
=
2
sin
θ
v
t
=
4
+
2
cot
θ
8
=
2
sin
θ
v
v
=
8
2
sin
θ
+
cos
θ
v
=
8
2
sin
θ
+
cos
θ
For minimum value of speed,
d
v
d
θ
=
0
d
v
d
θ
= 0
,
⇒
d
v
d
θ
=

8
x
(
2
cos
θ

sin
θ
)
(
2
sin
θ
+
cos
θ
)
2
=
0
⇒
(
2
cos
θ

sin
θ
)
=
0
⇒
tan
θ
=
2
⇒
d
v
d
θ
=

8
x
(
2
cos
θ

sin
θ
)
(
2
sin
θ
+
cos
θ
)
2
=
0
⇒
(
2
cos
θ

sin
θ
)
=
0
⇒
tan
θ
=
2
In order to evaluate the expression of velocity with trigonometric ratios, we take the help of right angle triangle as shown in the figure, which is consistent with the above result.
From the triangle, defining angle “θ”, we have :
sin
θ
=
2
√
5
sin
θ
=
2
√
5
and
cos
θ
=
1
√
5
cos
θ
=
1
√
5
The minimum velocity is :
v
=
8
2
x
2
√
5
+
1
√
5
=
8
√
5
=
3.57
m
/
s
v
=
8
2
x
2
√
5
+
1
√
5
=
8
√
5
=
3.57
m
/
s