Train a leaves new york at 9am eastern time on monday, headed for los angeles at a constant rate of 40 miles per hour. on the same 3,000-mile stretch of track, train b leaves los angeles at noon eastern time on monday, traveling to new york at its constant rate of 60 miles per hour, with one exception: train b is delayed for exactly 2 hours in las vegas (approximately 200 miles from los angeles) due to track maintenance. assuming the time it takes for train b to decelerate and re-accelerate when stopping and resuming in las vegas is negligible, at what time (eastern time) will the two trains meet?

Question

Julien Alvarez

Mathematics

19 November, 02:45

Train a leaves new york at 9am eastern time on monday, headed for los angeles at a constant rate of 40 miles per hour. on the same 3,000-mile stretch of track, train b leaves los angeles at noon eastern time on monday, traveling to new york at its constant rate of 60 miles per hour, with one exception: train b is delayed for exactly 2 hours in las vegas (approximately 200 miles from los angeles) due to track maintenance. assuming the time it takes for train b to decelerate and re-accelerate when stopping and resuming in las vegas is negligible, at what time (eastern time) will the two trains meet?

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Answers (1)

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Jan Hampton · Answer 1

We can assume that B leaves Los Angeles not at noon but at 2pm Eastern Time, so 5 hours later than A leaves New York.

In 5 hours A covers 5*40=200 miles. So, A and B together should cover 3,000-200=2,800 miles.

Combined rate = 40 + 60 = 100 miles per hour.

To cover 2,800 miles they'll need 2,800/100=28 hours.

2pm Eastern Time on Monday + 28 hours = 6pm Eastern Time on Tuesday.

Answer: 6:00pm on Tuesday