SOLUTION: A motorboat left a harbor and traveled to an island at an average rate of 11 knots. The average speed on the return trip was 10 knots. If the total trip took 10.5 hours, how far is
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Question 508301: A motorboat left a harbor and traveled to an island at an average rate of 11 knots. The average speed on the return trip was 10 knots. If the total trip took 10.5 hours, how far is the harbor from the island? Found 2 solutions by Alan3354, oberobic:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! The avg speed for the round trip is
2*11*10/(11+10) = 220/21 knots
d = r*t = (220/21)*10.5
d = 110 nautical miles round trip
distance = 55 NM
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55/10 + 55/11 = 10.5 hours as a check
You can put this solution on YOUR website! d = r*t is the basic distance equation, where d = distance, r= rate, and t = time
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We have two trips of equal distance here, but we don't know the distance.
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x = time going to the island
and
11 = rate of speed going to the island
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We know the two trips took 10.5 hr in total
so
10.5-x = time coming back from the island
and
10 = rate of speed coming back
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d=d
so
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11x = 10*(10.5-x)
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11x = 105 -10x
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21x = 105
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x = 105/21
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x = 5 hr
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d = 11 * 5 = 55
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Check using the return trip.
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10.5 -5 = 5.5
10 * 5.5 = 55
Correct.
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Answer: The island is 55 knots from the harbor.
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Note that the unit of measurement 'knots' can mean nautical miles and nautical miles per hr.
In this case, both meanings are knot are used.
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Done.
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