A runs 1 2/3 times as fast as B. If A gives B a start of 80 m, how far from A's start should the winning post be so that they finish together?

Problem restatement
A's speed is 5/3 of B's. B starts 80 m ahead. Find the race length (from A's start) so that both finish simultaneously.

Given data

• v_A = (5/3) v_B (i.e., A is 1 2/3 times as fast).
• B's head start = 80 m.

Concept/Approach
Equal finishing times ⇒ distance/speed equal for both, with B's distance shorter by the head start.

Step-by-step calculation
Let race length from A's start be D. Time equality: D ÷ v_A = (D − 80) ÷ v_B Substitute v_A = (5/3)v_B ⇒ D ÷ [(5/3)v_B] = (D − 80) ÷ v_B (3/5)D = D − 80 ⇒ (2/5)D = 80 ⇒ D = 200 m

Verification/Alternative
Times: A takes 200 ÷ (5/3 v_B) = 120 ÷ v_B; B takes (200 − 80) ÷ v_B = 120 ÷ v_B (same).

Common pitfalls

• Adding the head start to the race length instead of subtracting from B's distance.

Final Answer
200 m