A runner starts from a point at 6:00 a.m. with a speed of 8 km/h. Another racer starts from the same point at 8:30 a.m. in the same direction with a speed of 10 km/h. At what time of the day (in p.m.) will the second racer overtake the first runner?

Introduction / Context:
This question is a classic catch-up problem in which one runner starts earlier at a slower speed, and a second racer starts later but runs faster along the same line and in the same direction. The key requirement is to determine when the faster racer catches up with the slower runner. This situation is common in relative speed questions and tests your understanding of head start and relative speed concepts.

Given Data / Assumptions:
- First runner starts at 6:00 a.m. with speed 8 km/h.
- Second racer starts at 8:30 a.m. with speed 10 km/h.
- Both move along the same straight route in the same direction at constant speeds.
- We need the clock time (in p.m.) when the second racer overtakes the first runner.

Concept / Approach:
Because the first runner starts earlier, he gains a head start distance by the time the second racer begins. After 8:30 a.m., the difference between their positions changes at the rate of the relative speed (difference of speeds) since both are moving in the same direction. The catch-up time from 8:30 a.m. can be found by dividing the head start distance by the relative speed. Finally, we add this catch-up time to 8:30 a.m. to obtain the overtaking time on the clock.

Step-by-Step Solution:
Step 1: Calculate the head start gained by the first runner before 8:30 a.m.Time from 6:00 a.m. to 8:30 a.m. = 2.5 hours.Distance covered by first runner in 2.5 hours = 8 * 2.5 = 20 km.Step 2: Compute the relative speed after 8:30 a.m.Speed of second racer = 10 km/h.Speed of first runner = 8 km/h.Relative speed (closing speed) = 10 - 8 = 2 km/h.Step 3: Time taken for the second racer to make up the 20 km head start at 2 km/h.Catch-up time = head start distance / relative speed = 20 / 2 = 10 hours.Step 4: Add this time to the second racer’s start time.Second racer starts at 8:30 a.m., so overtaking time = 8:30 a.m. + 10 hours = 6:30 p.m.

Verification / Alternative check:
Check distances at 6:30 p.m. From 6:00 a.m. to 6:30 p.m. is 12.5 hours. Distance of first runner = 8 * 12.5 = 100 km. From 8:30 a.m. to 6:30 p.m. is 10 hours. Distance of second racer = 10 * 10 = 100 km. Since both have covered 100 km by that time, they are at the same position, confirming that the overtaking happens at 6:30 p.m.

Why Other Options Are Wrong:
- 4:00 p.m. would give the second racer only 7.5 hours of running, covering 75 km, while the first runner would have covered 8 * 10 = 80 km, so they would not yet meet.
- 5:30 p.m. similarly yields unequal distances that do not match at the same instant.
- 8:00 p.m. is too late; by then, the faster racer would have passed the first runner earlier, not at that time.

Common Pitfalls:
Some students forget to convert the 2.5 hours head start correctly or mistakenly add speeds instead of subtracting them for same-direction motion. Another error is to compute the catch-up time from 6:00 a.m. instead of 8:30 a.m., which misaligns the time-line. Carefully distinguishing between the head start period and the catch-up period helps avoid these issues.

Final Answer:
The second racer will overtake the first runner at 6:30 p.m.