A policeman chases a thief on a straight road. The policeman catches the thief in 15 minutes if he drives with a speed of 60 km/h and catches the thief in 0.5 hours if he drives with a speed 45 km/h. What should be his driving speed (in km/h) to catch the thief in just 10 minutes?
Policeman Thief Chase: Calculating Driving Speed
This problem involves the relationship between distance, speed, and time. We need to find the policeman's speed required to catch a thief in a specific time (10 minutes). We are given two scenarios where the policeman catches the thief at different speeds and times. We can use this information to determine the initial distance between the policeman and the thief and the speed at which the thief is traveling.
Let $D_0$ be the initial distance between the policeman and the thief (in km).
Let $s_t$ be the speed of the thief (in km/h).
Let $s_p$ be the speed of the policeman (in km/h).
Let $t$ be the time taken to catch the thief (in hours).
The key concept here is relative speed. The policeman needs to cover the initial distance $D_0$ relative to the thief. The relative speed of the policeman with respect to the thief is $(s_p - s_t)$.
The formula relating distance, relative speed, and time is:
$$ D_0 = (s_p - s_t) \times t $$Calculating Thief's Speed and Initial Distance
We can use the information from the two given scenarios to find $s_t$ and $D_0$. First, let's convert the times given in minutes to hours:
- Scenario 1: Time $t_1 = 15 \text{ minutes} = \frac{15}{60} \text{ hours} = 0.25 \text{ hours}$. Policeman's speed $s_{p1} = 60 \text{ km/h}$.
- Scenario 2: Time $t_2 = 0.5 \text{ hours}$. Policeman's speed $s_{p2} = 45 \text{ km/h}$.
Now, let's set up equations for both scenarios:
Scenario 1:
$$ D_0 = (s_{p1} - s_t) \times t_1 $$ $$ D_0 = (60 - s_t) \times 0.25 \quad \quad (1) $$Scenario 2:
$$ D_0 = (s_{p2} - s_t) \times t_2 $$ $$ D_0 = (45 - s_t) \times 0.5 \quad \quad (2) $$Since the initial distance $D_0$ is the same in both cases, we can equate the right-hand sides of equations (1) and (2):
$$ (60 - s_t) \times 0.25 = (45 - s_t) \times 0.5 $$Now, let's solve for $s_t$:
$$ 15 - 0.25 s_t = 22.5 - 0.5 s_t $$Rearrange the terms to group $s_t$ terms on one side and constants on the other:
$$ 0.5 s_t - 0.25 s_t = 22.5 - 15 $$ $$ 0.25 s_t = 7.5 $$ $$ s_t = \frac{7.5}{0.25} $$ $$ s_t = 30 \text{ km/h} $$So, the thief's speed is 30 km/h. Now we can find the initial distance $D_0$ by substituting $s_t = 30$ into either equation (1) or (2).
Using equation (1):
$$ D_0 = (60 - 30) \times 0.25 $$ $$ D_0 = 30 \times 0.25 $$ $$ D_0 = 7.5 \text{ km} $$The initial distance between the policeman and the thief is 7.5 km.
Determining Policeman's Speed for 10 Minutes
We need to find the policeman's speed ($s_{p3}$) required to catch the thief in $t_3 = 10$ minutes. First, convert this time to hours:
$$ t_3 = 10 \text{ minutes} = \frac{10}{60} \text{ hours} = \frac{1}{6} \text{ hours} $$We use the same relative speed formula, with the known $D_0 = 7.5$ km, $s_t = 30$ km/h, and $t_3 = \frac{1}{6}$ hours:
$$ D_0 = (s_{p3} - s_t) \times t_3 $$ $$ 7.5 = (s_{p3} - 30) \times \frac{1}{6} $$To solve for $s_{p3}$, first multiply both sides by 6:
$$ 7.5 \times 6 = s_{p3} - 30 $$ $$ 45 = s_{p3} - 30 $$Now, add 30 to both sides:
$$ s_{p3} = 45 + 30 $$ $$ s_{p3} = 75 \text{ km/h} $$Therefore, the policeman should drive at a speed of 75 km/h to catch the thief in 10 minutes.
Similar Questions
- A policeman spots a thief from a distance of 100 metres. When he starts running after the thief, the thief starts running as well. How far will the thief have to run before being overtaken if his speed is 8 km/h and the policeman's speed is 10 km/h?
- A policeman follows a thief who is 400 m ahead of him. If they run at the speed of 7 km/h and 6 km/h, respectively, what distance does the policeman run to catch up with the thief?
- A thief is noticed by a policeman from a distance of 800 m. The thief starts running and the policeman chases him. The thief and the policeman run at speeds of 20 km/h and 25 km/h, respectively. What is the distance between them after 9 minutes?
Important Questions from Relative Speed
- Ankur jogs on a circular track X at $1 \frac{2}{3}$ m/s and Mahima jogs on track Y at $2 \frac{1}{12}$ m/s. The total length of Tracks X and Y is 0.325 km. Ankur makes 9 rounds of track X in the same time Mahima makes 5 rounds of track Y. The time taken (in seconds) by Mahima to complete one round of track X is:
- Two places, A and B are 3710 kms from each other. A train leaves from A to B at the same time another train leaves B to A. The two trains meet at the end of 35 hours. If the train travelling from A to B travels 35 kmph faster than the other. Find the speed of the faster train. (In kmph)
- A truck, 1575 metres long is running at a speed of 109 km/hr. How many seconds will it take to cross a 1475 metres long train running in the same direction at a speed of 79 km/hr?
- A truck, 1600 metres long is running at a speed of 112 km/hr. How many seconds will it take to cross a 1500 metres long train running in the same direction at a speed of 82 km/hr?
- Two friends 10 km apart start running towards each other at speeds of 10 km/hr and 14 km/hr respectively. After how much time will they meet each other?
