In a 1 km race, A beats B by 18 m or 9 s. Find A's time over the 1000 m course (in seconds).

Introduction / Context:Race problems often give two equivalent margins: a distance lead or a time lead. Using either description, we can infer one runner's speed and then compute the finish time of the other runner over a fixed distance.

Given Data / Assumptions:

• Total race length = 1000 m.
• A beats B by 18 m or by 9 seconds (these describe the same finish margin).
• Uniform speeds throughout the race.

Concept / Approach:At A's finish, B is 18 m short. The “18 m equals 9 s” statement pins down B's speed. From B's speed, compute B's time for 1000 m, then subtract the 9 s gap to get A's time.

Step-by-Step Solution: B's speed = 18 m / 9 s = 2 m/s B's time for 1000 m = 1000 / 2 = 500 s Since A finishes 9 s quicker, A's time = 500 − 9 = 491 s

Verification / Alternative check:At t = 491 s, A completes 1000 m. In 491 s at 2 m/s, B covers 982 m, i.e., 18 m behind, exactly matching the given margin.

Why Other Options Are Wrong:500 s is B's time, not A's. 391 s and 591 s are inconsistent with the 18 m ↔ 9 s conversion and total distance of 1000 m.

Common Pitfalls:Using A's instead of B's speed from the 18 m/9 s linkage; mixing up who is faster; converting meters to seconds incorrectly.

Final Answer:491 s