Father–son “lost and found” with a shuttle dog: A man and his son get separated at a fair. The father realizes this after 20 minutes. The son turns back toward home at 20 m/min, and at that same moment the father moves toward the son at 40 m/min. A dog runs continuously between them at 60 m/min, shuttling back and forth, until father and son meet. What total distance does the dog cover <em>while running toward the son</em> (i.e., in the father→son legs)?

Introduction / Context:
This classic shuttle-dog puzzle uses relative speed to find how long two people take to meet. The dog’s total running time equals the meeting time because it runs continuously. The “toward the son” distance is the sum of only those shuttle legs from father to son.

Given Data / Assumptions:

• Upon realizing the separation, we assume the son had gone ahead for 20 min at the same 20 m/min and immediately turns back.
• At t = 0 (realization moment), separation = 20 * 20 = 400 m.
• Father speed toward son = 40 m/min; son’s return speed = 20 m/min.
• Dog runs at 60 m/min, instantly reversing at each turnaround, until meeting.

Concept / Approach:
The father and son meet after time T = separation / (sum of approach speeds) = 400 / (40 + 20). The dog runs for T minutes in total. Because shuttle legs are brief and frequent, the sum of all legs toward son equals half of the dog’s total distance when the approach speeds are in a constant ratio and the dog alternates symmetrically between them; here, symmetry holds due to constant speeds and instantaneous turnarounds.

Step-by-Step Solution:

Verification / Alternative check:
Direct summation via geometric sequence for shuttle legs also converges to equal halves under instantaneous reversals and constant speeds.

Why Other Options Are Wrong:
Values like 800 m or 1000 m assume the dog runs for 20+ minutes or misreads the initial separation. 848 m or 1675 m have no consistent derivation from the given constants.

Common Pitfalls:
Mixing “to-and-fro total” with “one-direction subtotal.” With alternating equal-time partition at constant shuttle speed and symmetric endpoints, the directional subtotal equals half the total; with the final approach landing exactly at the meeting, halves balance.

Final Answer:
400 m