Policeman Thief Chase: Understanding the Problem

This problem involves a classic chase scenario where we need to determine the distance at which a policeman catches a thief. We are given their speeds and the fact that the policeman starts later than the thief.

Analyzing Speeds and Time Difference

• Policeman's speed, denoted as $v_p$, is 18 km/h.
• Thief's speed, denoted as $v_t$, is 6 km/h.
• The policeman starts 10 minutes late. We need to convert this time difference into hours for consistency with the speed units:
  Time difference ($\Delta t$) = 10 minutes = $\frac{10}{60}$ hours = $\frac{1}{6}$ hour.

Calculating the Thief's Head Start

Because the policeman started late, the thief had a head start. We can calculate the distance the thief covered during this time:

Head Start Distance ($D_{headstart}$) = Thief's speed $\times$ Time difference

$D_{headstart} = v_t \times \Delta t

$D_{headstart} = 6 \text{ km/h} \times \frac{1}{6} \text{ h}

$D_{headstart} = 1 \text{ km}

So, when the policeman begins the chase, the thief is already 1 km ahead.

Applying the Concept of Relative Speed

To find out how quickly the policeman closes the gap, we use the concept of relative speed. Since the policeman is chasing the thief, the relative speed is the difference between their speeds:

Relative Speed ($v_{rel}$) = Policeman's speed - Thief's speed

$v_{rel} = v_p - v_t

$v_{rel} = 18 \text{ km/h} - 6 \text{ km/h}

$v_{rel} = 12 \text{ km/h}

This means the distance between the policeman and the thief decreases at a rate of 12 km per hour.

Calculating the Time to Catch the Thief

Now, we can calculate the time it takes for the policeman to cover the 1 km head start distance at the relative speed:

Time to catch ($t_{catch}$) = Head Start Distance / Relative Speed

$t_{catch} = \frac{D_{headstart}}{v_{rel}}

$t_{catch} = \frac{1 \text{ km}}{12 \text{ km/h}}

$t_{catch} = \frac{1}{12} \text{ hour}

Determining the Catch Distance

Finally, to find the distance from the starting point where the policeman catches the thief, we calculate how far the policeman travels during this time ($t_{catch}$):

Catch Distance ($D_{catch}$) = Policeman's speed $\times$ Time to catch

$D_{catch} = v_p \times t_{catch}

$D_{catch} = 18 \text{ km/h} \times \frac{1}{12} \text{ h}

$D_{catch} = \frac{18}{12} \text{ km}

$D_{catch} = \frac{3}{2} \text{ km}

$D_{catch} = 1.5 \text{ km}

Alternatively, we can calculate the total distance traveled by the thief. The thief travels for $\frac{1}{6}$ hour initially plus $\frac{1}{12}$ hour after the policeman starts:

Total time thief travels = $\frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$ hour.

Total distance thief travels = $v_t \times (\text{Total time thief travels}) = 6 \text{ km/h} \times \frac{1}{4} \text{ h} = \frac{6}{4} \text{ km} = 1.5 \text{ km}$.

Both methods confirm that the policeman catches the thief at a distance of 1.5 km.