Solving the Police and Thief Speed Problem

This problem involves calculating speeds based on relative motion scenarios. We need to find the speed of the thief using the information provided about chase and meeting times.

Setting Up Relative Speed Equations

Let $P$ be the speed of the police officer in meters per minute (m/min) and $T$ be the speed of the thief in meters per minute (m/min).

The initial distance between them is 500 meters.

Case 1: Running in the Same Direction (Chase)

• The police officer catches the thief in 10 minutes.
• In this case, the relative speed is the difference between their speeds: $P - T$.
• Distance = Relative Speed $\times$ Time
• $500 \text{ m} = (P - T) \times 10 \text{ min}$
• This gives us the first equation: $P - T = \frac{500}{10} = 50$ m/min.

Case 2: Running Towards Each Other (Meeting)

• They meet in 2 minutes when running towards each other.
• The relative speed is the sum of their speeds: $P + T$.
• Distance = Relative Speed $\times$ Time
• $500 \text{ m} = (P + T) \times 2 \text{ min}$
• This gives us the second equation: $P + T = \frac{500}{2} = 250$ m/min.

Calculating the Thief's Speed

We now have a system of two linear equations:

1. $P - T = 50$
2. $P + T = 250$

To find the speed of the thief ($T$), we can subtract the first equation from the second:

$(P + T) - (P - T) = 250 - 50$

$P + T - P + T = 200$

$2T = 200$

$T = \frac{200}{2}$

$T = 100$ m/min.

Alternatively, we can first find the police officer's speed ($P$) by adding the two equations:

$(P - T) + (P + T) = 50 + 250$

$2P = 300$

$P = 150$ m/min.

Substitute $P = 150$ into the second equation ($P + T = 250$):

$150 + T = 250$

$T = 250 - 150$

$T = 100$ m/min.

The speed of the thief is 100 meters per minute.