Understanding the Speed and Distance Problem

This question involves concepts of speed, distance, and time. We have two individuals, a thief and a policeman, moving at different speeds, with the policeman starting later. We need to find the policeman's speed and the meeting point from the crime spot.

Analyzing the Initial Lead of the Thief

The thief starts running at a speed of 15 km/h. The policeman arrives 4 minutes later and immediately begins chasing. This means the thief gets a head start of 4 minutes.

• Thief's speed ($S_T$): $15$ km/h
• Policeman's delay: $4$ minutes
• Convert delay to hours: $4 \text{ minutes} = \frac{4}{60} \text{ hours} = \frac{1}{15} \text{ hours}$

During this 4-minute head start, the thief covers a certain distance:

Distance covered by thief during head start = $S_T \times \text{Time}$

Distance = $15 \text{ km/h} \times \frac{1}{15} \text{ hours} = 1$ km.

So, when the policeman starts chasing, the thief is already 1 km away from the crime spot.

Analyzing the Chase Phase

The policeman chases the thief for 16 minutes, and at that point, there is still a gap of 200 m between them.

• Chase time: $16$ minutes
• Convert chase time to hours: $16 \text{ minutes} = \frac{16}{60} \text{ hours} = \frac{4}{15} \text{ hours}$
• Remaining gap: $200$ m
• Convert remaining gap to km: $200 \text{ m} = \frac{200}{1000} \text{ km} = 0.2$ km

Distance Covered During Chase Time

In these 16 minutes, the thief continues running at 15 km/h.

Distance covered by thief in 16 minutes = $S_T \times \text{Time}$

Distance = $15 \text{ km/h} \times \frac{4}{15} \text{ hours} = 4$ km.

The total distance the thief is from the crime spot after the 16 minutes of policeman chasing is the initial head start distance plus the distance covered during the chase time:

Total distance of thief from spot = $1 \text{ km (head start)} + 4 \text{ km (during chase)} = 5$ km.

At this point (after 16 minutes of chasing), the policeman is 200 m (0.2 km) behind the thief. This means the policeman has covered a distance that is 0.2 km less than the thief's total distance from the spot.

Distance covered by policeman in 16 minutes = Total distance of thief from spot - Remaining gap

Distance covered by policeman = $5 \text{ km} - 0.2 \text{ km} = 4.8$ km.

Calculating the Policeman's Speed

The policeman covered 4.8 km in 16 minutes ($4/15$ hours). We can now calculate the policeman's speed ($S_P$).

$S_P = \frac{\text{Distance covered by policeman}}{\text{Time taken by policeman}}$

$S_P = \frac{4.8 \text{ km}}{\frac{4}{15} \text{ hours}}$

$S_P = 4.8 \times \frac{15}{4}$ km/h

$S_P = \frac{48}{10} \times \frac{15}{4}$ km/h

$S_P = \frac{12}{10} \times 15$ km/h

$S_P = 1.2 \times 15$ km/h

$S_P = 18$ km/h.

The speed of the policeman is 18 km/h.

Calculating the Catch-Up Distance from the Spot

To find the distance from the crime spot where the policeman catches the thief, we first need to find the time it takes for the policeman to catch up. Let $T$ be the time (in hours) the policeman chases the thief until he catches up. During this time $T$, the policeman covers a distance $S_P \times T$.

In the same time $T$, the thief covers a distance $S_T \times T$. However, the thief had a 1 km head start.

When the policeman catches up, the distance covered by the policeman from the spot equals the total distance covered by the thief from the spot.

Distance covered by policeman = Distance covered by thief (head start) + Distance covered by thief (during chase T)

$S_P \times T = 1 \text{ km} + S_T \times T$

Substitute the speeds we know:

$18 \times T = 1 + 15 \times T$

Now, solve for $T$:

$18T - 15T = 1$

$3T = 1$

$T = \frac{1}{3}$ hours.

This means the policeman catches the thief $\frac{1}{3}$ hours after he started chasing. $\frac{1}{3}$ hours is equal to $\frac{1}{3} \times 60 = 20$ minutes.

Now, calculate the distance from the crime spot where this happens. We can use either the policeman's distance or the thief's total distance.

Using policeman's distance:

Distance = $S_P \times T$

Distance = $18 \text{ km/h} \times \frac{1}{3} \text{ hours} = 6$ km.

Using thief's total distance:

Total distance = Distance during head start + Distance during chase T

Total distance = $1 \text{ km} + (S_T \times T)$

Total distance = $1 \text{ km} + (15 \text{ km/h} \times \frac{1}{3} \text{ hours})$

Total distance = $1 \text{ km} + 5 \text{ km} = 6$ km.

The policeman catches up to the thief at a distance of 6 km from the spot of the crime.

Summary of Results

Thus, the policeman catches up the thief at a distance of 6 km from the spot of the crime, and the speed of the policeman is 18 km/h.

Revision Table: Speed, Distance, Time Formulas

Additional Information: Relative Speed in Chasing Problems

In problems where one object chases another, the concept of relative speed is very useful. When the objects are moving in the same direction, their relative speed is the difference between their speeds ($S_{relative} = S_1 - S_2$).

In this problem, the thief has a 1 km head start when the policeman begins. The policeman needs to cover this initial 1 km gap using the relative speed difference between himself and the thief.

• Thief's speed ($S_T$): 15 km/h
• Policeman's speed ($S_P$): 18 km/h (calculated earlier)
• Relative speed ($S_P - S_T$): $18 - 15 = 3$ km/h.

The time taken to cover the initial 1 km gap using relative speed is:

Time = $\frac{\text{Initial Gap}}{\text{Relative Speed}}$

Time = $\frac{1 \text{ km}}{3 \text{ km/h}} = \frac{1}{3}$ hours.

This is the same time $T$ we calculated earlier, representing the time from when the policeman starts chasing until he catches up. This confirms our time calculation.

The distance from the crime spot where the catch-up occurs is the distance the policeman covers in this time:

Distance = Policeman's Speed $\times$ Time

Distance = $18 \text{ km/h} \times \frac{1}{3} \text{ hours} = 6$ km.

This confirms the distance calculation as well. Using relative speed can sometimes simplify the approach to these types of speed, distance, and time problems.