Problem: A police car chases a robber who has a 5-minute head start. The police car goes 20 mph faster than the robber. If the police car catches up to the robber in 30 minutes, what is the speed of each vehicle?

Solution:

Let’s start by defining our variables. Let’s call the speed of the robber “r” and the speed of the police car “p”. We know that the police car goes 20 mph faster than the robber, so we can write:

p = r + 20

We also know that the police car catches up to the robber in 30 minutes. Let’s call the distance the robber travels “d”. Since the police car catches up to the robber, we know that they both travel the same distance. The robber has a 5-minute head start, so he travels for 30 + 5 = 35 minutes. We can write:

d = r * 35 (distance traveled by the robber) d = p * 30 (distance traveled by the police car)

Since they both travel the same distance, we can set these two equations equal to each other:

r * 35 = p * 30

Now we can substitute the expression we found for “p” into this equation:

r * 35 = (r + 20) * 30

Simplifying this equation, we get:

35r = 30r + 600

5r = 600

r = 120

So the speed of the robber is 120 mph. We can use the expression we found for “p” to find the speed of the police car:

p = r + 20 = 120 + 20 = 140

So the speed of the police car is 140 mph.

Therefore, the speed of the robber is 120 mph and the speed of the police car is 140 mph.