Solving the Journey Catch-up Problem

This problem involves analyzing the movements of two people, Suraj and Rahul, starting at different times from the same point and traveling at different speeds. We need to find the time when Rahul catches up to Suraj.

Understanding Suraj's Journey

Suraj starts at 2 p.m. from point Q. His journey has two phases with different speeds:

• Phase 1: For the first 3 hours (from 2 p.m. to 5 p.m.), Suraj travels at a speed of 40 km/hr.
• Phase 2: After the first 3 hours (from 5 p.m. onwards), Suraj travels at a speed of 30 km/hr.

Analyzing Rahul's Pursuit

Rahul starts his journey from the same point Q at 6 p.m. He travels towards Suraj at a constant speed of 80 km/hr.

Calculating Distances at Key Times

To find when Rahul catches Suraj, we first need to determine Suraj's position at 6 p.m., which is when Rahul starts.

• Distance covered by Suraj in the first 3 hours (2 p.m. to 5 p.m.):
  Distance = Speed × Time
  Distance = \(40 \text{ km/hr} \times 3 \text{ hours} = 120 \text{ km}\)
• Time elapsed for Suraj from 5 p.m. to 6 p.m. is 1 hour. During this hour, he travels at 30 km/hr.
• Distance covered by Suraj from 5 p.m. to 6 p.m.:
  Distance = Speed × Time
  Distance = \(30 \text{ km/hr} \times 1 \text{ hour} = 30 \text{ km}\)
• Total distance of Suraj from point Q at 6 p.m.:
  Total Distance = Distance (2-5 p.m.) + Distance (5-6 p.m.)
  Total Distance = \(120 \text{ km} + 30 \text{ km} = 150 \text{ km}\)

So, at 6 p.m., Suraj is 150 km away from point Q, and Rahul is at point Q.

Setting up the Catch-up Equation

Let \(t\) be the time in hours after 6 p.m. when Rahul catches Suraj. At the moment Rahul catches Suraj, they will both be at the same distance from point Q.

• Distance covered by Rahul in \(t\) hours (starting from Q at 6 p.m.):
  Distance by Rahul = Speed × Time
  Distance by Rahul = \(80 \text{ km/hr} \times t \text{ hours} = 80t \text{ km}\)
• Distance covered by Suraj in \(t\) hours (starting from his position at 6 p.m.):
  At 6 p.m., Suraj was 150 km from Q. After 6 p.m., Suraj continues to travel at 30 km/hr.
  Distance covered by Suraj after 6 p.m. in \(t\) hours = \(30 \text{ km/hr} \times t \text{ hours} = 30t \text{ km}\).
  Total distance of Suraj from Q at time \(t\) hours after 6 p.m. = Initial Distance + Distance covered after 6 p.m.
  Total Distance by Suraj = \(150 \text{ km} + 30t \text{ km}\)

When Rahul catches Suraj, their distances from Q are equal:

\(80t = 150 + 30t\)

Solving for Catch-up Time

Now we solve the equation for \(t\):

• Subtract \(30t\) from both sides of the equation:
  \(80t - 30t = 150\)
• Simplify:
  \(50t = 150\)
• Divide by 50:
  \(t = \frac{150}{50}\)
  \(t = 3 \text{ hours}\)

This means Rahul catches Suraj 3 hours after 6 p.m.

Determining the Final Catch-up Time

Rahul started at 6 p.m. He catches Suraj 3 hours later.

Catch-up time = 6 p.m. + 3 hours = 9 p.m.

Therefore, Rahul will catch Suraj at 9:00 p.m.

Summary of Calculations

The calculated time of 9:00 p.m. matches one of the given options.

Revision Table: Journey Problem Concepts

Additional Information: Distance-Time Problems

Distance-time problems are common in mathematics and physics. They often involve objects moving at constant speeds or speeds that change over segments of the journey. The key is to carefully track the position of each object at different points in time and set up equations based on when and where they meet or when one overtakes the other.

For problems where one person chases another, calculating the relative speed can simplify the process. The relative speed is the rate at which the distance between the two people closes. The time taken to catch up is the initial distance between them divided by the relative speed.

In this specific problem, at 6 p.m., the initial distance between Rahul and Suraj was 150 km. Since they are moving in the same direction (Rahul chasing Suraj), the relative speed is Rahul's speed minus Suraj's speed (80 km/hr - 30 km/hr = 50 km/hr). The time to catch up is \(150 \text{ km} / 50 \text{ km/hr} = 3\) hours. Since Rahul starts at 6 p.m., the catch-up time is 6 p.m. + 3 hours = 9 p.m.