Average speed over unequal segments A man covers the first 30 km at 6 km/h and the remaining 40 km in 5 hours. What is his average speed for the 70 km journey?
Correct Answer: 7 km/h
Introduction / Context:Average speed over multiple legs equals total distance divided by total time, not the average of individual speeds. When legs have different durations, using total distance / total time ensures correct weighting.
Given Data / Assumptions:
- Leg 1: 30 km at 6 km/h.
- Leg 2: 40 km in 5 h (so the speed on leg 2 is 8 km/h).
- Total distance = 70 km.
Concept / Approach:Compute time per leg, sum to total time, then divide total distance by total time.
Step-by-Step Solution:
Time for first 30 km = 30 / 6 = 5 hTime for last 40 km = 5 hTotal time = 5 + 5 = 10 hAverage speed = Total distance / Total time = 70 / 10 = 7 km/hVerification / Alternative check:Weighted-by-time average of speeds: (6 for 5 h and 8 for 5 h) ⇒ (6*5 + 8*5) / (5+5) = 70/10 = 7.
Why Other Options Are Wrong:8 km/h uses the last leg speed; 64/11 ≈ 5.82 and 7.5 km/h mis-handle weighting.
Common Pitfalls:Taking the arithmetic mean (6+8)/2 = 7 (which coincidentally equals the correct value only because the times are equal here; in general you must use total distance / total time).
Final Answer:7 km/h