Race timing with head-start: Runner A covers 440 m in 51 s, while runner B covers the same 440 m in 55 s. In a 440 m race, if B is given a 40 m head-start (so B needs to run only 400 m), by how many seconds will B finish ahead of A?

Introduction / Context:Head-start problems compare finishing times using constant speeds. We compute each runner’s time to complete their required distance and compare.

Given Data / Assumptions:

• A runs 440 m in 51 s.
• B runs 440 m in 55 s.
• Race length = 440 m; B receives a 40 m head-start, so B runs 400 m.
• Uniform speed throughout.

Concept / Approach:Time = distance / speed. Speeds are vA = 440/51 m/s and vB = 440/55 m/s. Compute tA and tB for their race distances and compare finishing times.

Step-by-Step Solution:

Verification / Alternative check:Proportionally, B’s pace is 55/51 slower than A’s, but the 40 m reduction at B’s speed equals 40 / (440/55) = 5 s saved, while B’s inherent slowness adds 4 s over 440 m, net 1 s ahead.

Why Other Options Are Wrong:

• 2/3/4 seconds: Overstate the time gap; calculations show exactly 1 s.
• None of these: Incorrect since 1 s matches an option.

Common Pitfalls:Using distance difference instead of time, or forgetting that B runs only 400 m, not 440 m.

Final Answer:1 seconds