A police car is located 40 feet to the side of a straight road.

Question

A police car is located 40 feet to the side of a straight road.

A red car is driving along the road in the direction of the police car and is 120 feet up the road from the location of the police car. The police radar reads that the distance between the police car and the red car is decreasing at a rate of 90 feet per second. How fast is the red car actually traveling along the road?

The actual speed (along the road) of the red car is  feet per second

Answer 1

the police are 40 feet from the road

the car approaches 120 feet away from where the police are perpendicular to the road. diagonal distance between the police and the car is decreasing at the rate of 90 feet per second

How fast is the car traveling?

this is a "related rate" problem. Most calculus textbooks will have a a few pages on this topic in the derivative section, with specific examples.

construct a right triangle with sides 40 and 120. the hypotenuse squared = 40^2 + 120^2 = 1600 + 14400 = 16000

hypotenuse = sqr16000 = 40sqr10= direct distance from the red car to the police

hypotenuse squared = sum of squares of the other two sides

h^2 = a^2 + b^2

take the derivative with respect to time, h'=dh/dt, a'=da/dt, b'=db/dt

2hh' = 2aa' + 2bb'

hh'- = aa' + bb' where b' = 0= change of distance from the police to the road which is zero, b= 40, h'=90, a = 120, a' = the speed of the red car, h=40sqr10

plug in the values into hh'=aa'+bb'

40sqr10(90) = 120a' + (40)(0)

3600sqr10 = 120a'

a' = sqr10(3600)/120 =30sqr10

= 94.86 ft/sec

= about 95 feet per second

convert to mph using an on line converter calculator. there's several websites that do it for free.

or use 1 mile=5280 feet and 1 hour = 3600 seconds:

95 ft/sec = red car's speed

= 95(3600)/5280 mph = about 64.77 mph

= about 65 mph = the red car's speed