Related Rates Problem
Two cars, one going due south at the rate of 3 0 km/hr and the other going due east at the rate of 4 5 km/hr , are travelling toward the intersection of two roads.
In kph , at what rate are the cars approaching each other at the instant when the first car is 0 . 2 km and the second car is 0 . 1 5 km from the intersection?
Submit your answer as a positive number.
The answer is 51.
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1 solution
Let time (
t
be zero when both cars are at their stated distance from the intersection. When
t
=
0
,
D
=
0
.
2
0
2
+
0
.
1
5
2
=
0
.
2
5
From the diagram, since the 1st car is heading south (negative direction), its distance from the intersection (origin) can be written as:
y
=
0
.
2
0
−
3
0
t
The 2nd car's distance from the intersection is:
x
=
−
0
.
1
5
+
4
5
t
By the Pythagorean theorem, the "distance squared" between the two cars at any point in time (including negative time) is:
D
2
2
D
d
t
d
D
v
(
t
)
=
d
t
d
D
=
x
2
+
y
2
=
(
−
0
.
1
5
+
4
5
t
)
2
+
(
0
.
2
0
−
3
0
t
)
2
=
0
.
0
6
2
5
−
2
5
.
5
t
+
2
9
2
5
t
2
=
−
2
5
.
5
+
5
8
5
0
t
(implicit differentiation)
=
2
D
−
2
5
.
5
(set t=0)
=
2
(
0
.
2
5
)
−
2
5
.
5
=
−
5
1
k
m
/
h
The rate of change of displacement of the two cars is negative, but since the question asks at what rate are the cars approaching, the answer can also be stated as a positive quantity: 51 km/h