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Difficulty: HardRate, Time, and Distance Problems

An autonomous electric shuttle completed a 240-mile test run from Point A to Point B. For the first 120 miles, the shuttle traveled at a constant speed of vv miles per hour. For the remaining 120 miles, due to battery efficiency controls, the shuttle traveled at a reduced speed equal to 23v\frac{2}{3}v miles per hour. On the return trip along the exact same 240-mile route from Point B to Point A, the shuttle maintained a uniform speed of v+10v + 10 miles per hour. If the total time for the outbound trip from Point A to Point B was 2 hours longer than the total time for the return trip from Point B to Point A, what was the shuttle's average speed, in miles per hour, for the outbound trip from Point A to Point B?

  1. A
    32
  2. 40Answer
  3. C
    412341\frac{2}{3}
  4. D
    45
  5. E
    50

Answer

40 miles per hour
To find the average speed for the outbound trip, calculate the total time taken for both 120-mile legs. The first half takes 120v\frac{120}{v} hours, and the second half at 23v\frac{2}{3}v speed takes 12023v=180v\frac{120}{\frac{2}{3}v} = \frac{180}{v} hours, giving a total outbound time of 300v\frac{300}{v} hours. Setting up the time difference equation against the return trip gives 300v−240v+10=2\frac{300}{v} - \frac{240}{v + 10} = 2. Solving this quadratic equation yields v=50v = 50 mph. Substituting v=50v = 50 into the total outbound time formula gives 30050=6\frac{300}{50} = 6 hours. Finally, dividing the total distance of 240 miles by 6 hours gives an average speed of 40 miles per hour.

Step-by-Step Solution

1
Express the total outbound travel time in terms of vv.
Toutbound=120v+12023v=120v+180v=300vT_{\text{outbound}} = \frac{120}{v} + \frac{120}{\frac{2}{3}v} = \frac{120}{v} + \frac{180}{v} = \frac{300}{v} hours.
Time equals distance divided by speed for each 120-mile segment.
2
Express the return travel time and set up the time-difference equation.
300v−240v+10=2\frac{300}{v} - \frac{240}{v + 10} = 2.
The return trip covers 240 miles at speed v+10v + 10 mph and takes 2 hours less than the outbound trip.
3
Solve the algebraic equation for vv.
Simplifying 150v−120v+10=1\frac{150}{v} - \frac{120}{v + 10} = 1 leads to v2−20v−1500=0v^2 - 20v - 1500 = 0, giving positive root v=50v = 50.
Factoring (v−50)(v+30)=0(v - 50)(v + 30) = 0 yields v=50v = 50 mph since speed must be positive.
4
Calculate total outbound time and outbound average speed.
Total outbound time = 30050=6\frac{300}{50} = 6 hours; Average speed = 2406=40\frac{240}{6} = 40 mph.
Average speed is defined as total distance divided by total elapsed time.

Key Concept

Average Speed across Multi-Leg Trips (Total Distance / Total Time)
Estimated Time:2m 30s
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Topics

Rate, Time, and Distance Problems

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