Understanding the Policeman and Thief Chase Problem

This problem requires us to calculate the separation distance between a policeman and a thief after a certain amount of time. We are given their initial distance apart, their respective speeds, and the time duration for which they run. The key is to figure out how much closer the policeman gets to the thief during the chase.

Extracting Key Information

Let's list the important values provided in the question:

• Initial distance between the policeman and the thief: 300 m
• Speed of the thief ($V_t$): 10 km/h
• Speed of the policeman ($V_p$): 12 km/h
• Time duration ($t$): 6 minutes
• The objective is to find the final distance between them in meters after 6 minutes.

Essential Unit Conversions

To perform the calculations correctly, we must ensure all units are consistent. The speeds are in kilometers per hour (km/h), the time is in minutes, and the distances are in meters (m). We will convert the speeds to meters per second (m/s) and the time to seconds (s).

• Time Conversion: The time duration is 6 minutes. To convert minutes to seconds, we multiply by 60. $t = 6 \text{ minutes} = 6 \times 60 \text{ seconds} = 360 \text{ seconds}$
• Speed Conversion (km/h to m/s): To convert speed from km/h to m/s, we multiply by the factor $\frac{1000 \text{ m}}{3600 \text{ s}}$, which simplifies to $\frac{5}{18}$.
  • Thief's speed: $V_t = 10 \text{ km/h} = 10 \times \frac{5}{18} \text{ m/s} = \frac{50}{18} \text{ m/s} = \frac{25}{9} \text{ m/s}$
  • Policeman's speed: $V_p = 12 \text{ km/h} = 12 \times \frac{5}{18} \text{ m/s} = \frac{60}{18} \text{ m/s} = \frac{10}{3} \text{ m/s}$

Calculating the Relative Speed

Since the policeman is chasing the thief, they are both moving in the same direction. The speed at which the distance between them decreases is called the relative speed. It is calculated by subtracting the speed of the slower object (thief) from the speed of the faster object (policeman).

Relative Speed ($V_{rel}$) = Policeman's Speed ($V_p$) - Thief's Speed ($V_t$)

Let's calculate this relative speed:

To subtract the fractions, we need a common denominator, which is 9.

This relative speed of $\frac{5}{9}$ m/s indicates that the policeman is closing the gap between them by this amount every second.

Calculating the Distance Closed

We now need to find out how much distance the policeman has effectively covered relative to the thief during the 6 minutes (360 seconds).

Distance Closed = Relative Speed ($V_{rel}$) $\times$ Time ($t$)

Using the values we have:

Distance Closed = $\frac{5}{9} \text{ m/s} \times 360 \text{ s}$

Distance Closed = $5 \times \frac{360}{9}$ m

Distance Closed = $5 \times 40$ m

Distance Closed = $200$ m

This calculation shows that the policeman has reduced the initial distance by 200 meters in 6 minutes.

Determining the Final Distance

The final step is to calculate the remaining distance between the policeman and the thief after these 6 minutes.

Final Distance = Initial Distance - Distance Closed

Plugging in the values:

Final Distance = $300 \text{ m} - 200 \text{ m}$

Final Distance = $100 \text{ m}$

Thus, after 6 minutes, the policeman and the thief will be 100 meters apart.