Thief Chase Problem: Calculating Speed and Distance

This problem involves calculating the speed of a policeman and the distance from the crime spot where they catch a thief, based on their respective speeds, a delay in starting the chase, and a known gap after a certain time.

Setting Up Variables and Units

First, let's define the knowns and unknowns, ensuring consistent units. We'll primarily use kilometers (km) for distance and hours (h) for time.

• Thief's speed, $v_t = 13.5$ km/h.
• Policeman starts chasing after a delay, $t_{\text{delay}} = 8$ minutes.
• The chase duration observed is $t_{\text{chase}} = 28$ minutes.
• The gap between the thief and policeman after 28 minutes of chase is $ \text{gap} = 540$ m $= 0.54$ km.
• Unknowns: Policeman's speed ($v_p$) and the distance from the crime spot where the policeman catches the thief ($d_{\text{catch up}}$).

Let's convert the time durations into hours:

• $t_{\text{delay}} = 8 \text{ minutes} = \frac{8}{60} \text{ hours} = \frac{2}{15}$ hours.
• $t_{\text{chase}} = 28 \text{ minutes} = \frac{28}{60} \text{ hours} = \frac{7}{15}$ hours.

Thief Head Start: Calculating Initial Distance

The thief escapes before the policeman starts chasing. We need to find out how far ahead the thief is when the chase begins.

Distance covered by the thief during the policeman's delay:

$d_{\text{head start}} = v_t \times t_{\text{delay}}$

$d_{\text{head start}} = 13.5 \text{ km/h} \times \frac{2}{15} \text{ h}$

$d_{\text{head start}} = \frac{135}{10} \times \frac{2}{15} = \frac{27}{2} \times \frac{2}{15} = \frac{27}{15} = \frac{9}{5} = 1.8$ km.

So, when the policeman starts running, the thief is already 1.8 km ahead.

Policeman Speed Calculation: Using Chase Data

We know the situation after 28 minutes of chase. Let's calculate the positions of both the thief and the policeman relative to the crime spot at this moment.

Thief's total distance from the crime spot after 28 minutes of chase:

This includes the head start distance plus the distance covered during the 28 minutes of chase.

$d_{\text{thief, at 28min}} = d_{\text{head start}} + v_t \times t_{\text{chase}}$

$d_{\text{thief, at 28min}} = 1.8 \text{ km} + 13.5 \text{ km/h} \times \frac{7}{15} \text{ h}$

$d_{\text{thief, at 28min}} = 1.8 + \frac{135}{10} \times \frac{7}{15} = 1.8 + \frac{27}{2} \times \frac{7}{15} = 1.8 + \frac{9}{2} \times \frac{7}{5} = 1.8 + \frac{63}{10}$

$d_{\text{thief, at 28min}} = 1.8 + 6.3 = 8.1$ km.

Policeman's distance from the crime spot after 28 minutes of chase:

Let the policeman's speed be $v_p$. The distance covered by the policeman is:

$d_{\text{policeman, at 28min}} = v_p \times t_{\text{chase}} = v_p \times \frac{7}{15}$ km.

Using the gap to find the policeman's speed:

The gap is the difference between the thief's distance and the policeman's distance:

$d_{\text{thief, at 28min}} - d_{\text{policeman, at 28min}} = \text{gap}$

$8.1 \text{ km} - (v_p \times \frac{7}{15} \text{ km}) = 0.54 \text{ km}$

Rearranging the equation to solve for $v_p \times \frac{7}{15}$:

$v_p \times \frac{7}{15} = 8.1 - 0.54 = 7.56$ km.

Now, we can find $v_p$:

$v_p = 7.56 \times \frac{15}{7}$

$v_p = \frac{756}{100} \times \frac{15}{7} = \frac{108}{100} \times 15 = \frac{108 \times 3}{20} = \frac{27 \times 3}{5} = \frac{81}{5}$

$v_p = 16.2$ km/h.

So, the policeman ran at a speed of 16.2 km/h.

Catch-Up Time: Closing the Final Gap

Now we need to find out how much *more* time the policeman needs to catch the thief, given the 0.54 km gap and the relative speed.

The relative speed of the policeman with respect to the thief is:

$v_{\text{rel}} = v_p - v_t = 16.2 \text{ km/h} - 13.5 \text{ km/h} = 2.7$ km/h.

The time required to close the remaining gap of 0.54 km at this relative speed is:

$t_{\text{additional}} = \frac{\text{gap}}{v_{\text{rel}}}$

$t_{\text{additional}} = \frac{0.54 \text{ km}}{2.7 \text{ km/h}} = \frac{54}{270} \text{ h} = \frac{1}{5}$ h.

This additional time is $\frac{1}{5} \times 60 = 12$ minutes.

The total time the policeman spent chasing the thief until catching him is:

$T_{\text{policeman chase}} = t_{\text{chase}} + t_{\text{additional}}$

$T_{\text{policeman chase}} = \frac{7}{15} \text{ h} + \frac{1}{5} \text{ h} = \frac{7}{15} + \frac{3}{15} = \frac{10}{15} = \frac{2}{3}$ hours.

Catch-Up Distance Calculation: Total Distance Covered

Finally, we calculate the distance from the spot of the crime where the policeman catches the thief. This is the distance the policeman covers in the total time he was chasing.

$d_{\text{catch up}} = v_p \times T_{\text{policeman chase}}$

$d_{\text{catch up}} = 16.2 \text{ km/h} \times \frac{2}{3} \text{ h}$

$d_{\text{catch up}} = \frac{162}{10} \times \frac{2}{3} = \frac{81}{5} \times \frac{2}{3} = \frac{27}{5} \times 2 = \frac{54}{5}$

$d_{\text{catch up}} = 10.8$ km.

Therefore, the policeman would catch up with the thief at a distance of 10.8 km from the spot of the crime, and the policeman's speed was 16.2 km/h.