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Difficulty: Very hardRate, Time, and Distance Problems

A delivery drone flies in a straight line from Hub A to Hub B against a constant headwind of 1010 miles per hour, taking 44 hours to complete the flight. On the return flight from Hub B to Hub A along the exact same path, the wind direction reverses to become a tailwind, and its speed increases by 5050 percent. If the return flight takes 22 hours and the drone maintains a constant airspeed in still air throughout both flights, what was the drone's average speed, in miles per hour, for the entire round trip?

  1. A
    162316\frac{2}{3}
  2. B
    2020
  3. 331333\frac{1}{3}Answer
  4. D
    3535
  5. E
    371237\frac{1}{2}

Answer

The drone's average speed for the entire round trip is 331333\frac{1}{3} miles per hour.
The drone's still-air speed vv is determined by equating the outbound and return distance equations: 4(v−10)=2(v+15)4(v - 10) = 2(v + 15), giving v=35v = 35 miles per hour. The outbound ground speed is 2525 miles per hour over 100100 miles, and the return ground speed is 5050 miles per hour over 100100 miles. Dividing the total round-trip distance of 200200 miles by the total time of 66 hours yields an average speed of 2006=3313\frac{200}{6} = 33\frac{1}{3} miles per hour.

Step-by-Step Solution

1
Express the effective ground speeds and distance for both legs of the trip in terms of the drone's still-air speed vv.
Outbound headwind speed = 1010 mph, ground speed = v−10v - 10 mph, outbound distance D=4(v−10)D = 4(v - 10). Return tailwind speed = 10×(1+0.50)=1510 \times (1 + 0.50) = 15 mph, return ground speed = v+15v + 15 mph, return distance D=2(v+15)D = 2(v + 15).
Distance equals speed multiplied by time, and wind speeds adjust the effective ground speed depending on direction.
2
Equate the distance expressions to solve for the still-air speed vv.
4(v−10)=2(v+15)  ⟹  4v−40=2v+30  ⟹  2v=70  ⟹  v=354(v - 10) = 2(v + 15) \implies 4v - 40 = 2v + 30 \implies 2v = 70 \implies v = 35 mph.
The distance from Hub A to Hub B is identical in both directions.
3
Calculate the one-way distance DD and total round-trip distance.
One-way distance D=4(35−10)=100D = 4(35 - 10) = 100 miles. Total round-trip distance =100+100=200= 100 + 100 = 200 miles.
Average speed requires total distance covered across both legs.
4
Calculate the average speed for the entire round trip using Average Speed=Total DistanceTotal Time\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}.
Total time =4+2=6= 4 + 2 = 6 hours. Average speed =2006=1003=3313= \frac{200}{6} = \frac{100}{3} = 33\frac{1}{3} mph.
Average speed over multiple legs is defined as total distance divided by total elapsed time.

Key Concept

Average Speed for Multi-Leg Journeys with Wind Vector Effects
Estimated Time:2m 30s
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Topics

Rate, Time, and Distance Problems

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