A fire engine starts pumping water at 9:20 am at the rate of 800 gallons per minute. another fire engine, pumping at the rate of 1000 gallons per minute, starts at 9:30 am. at what time will the two engines have pumped the same number of gallons?

Question

31-12-2017
Physics

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A fire engine starts pumping water at 9:20 am at the rate of 800 gallons per minute. another fire engine, pumping at the rate of 1000 gallons per minute, starts at 9:30 am. at what time will the two engines have pumped the same number of gallons?

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AL2006 AL2006 · Answer 1 · 2017-12-31T06:15:46+01:00

When they both pump, Engine #2 gains quantity over Engine #1 at the rate of 200 gallons per minute.

BUT ... when Engine #2 started up at 9:30, Engine #1 was already

(800 x 10 minutes) = 8,000 gallons

in the lead. Engine #2 has to gain 8,000 gallons just to pull even.

Since Engine #2 gains 200 gallons per minute, it will take him

(8,000) / (200) = 40 minutes

to pull even with Engine #1.

That moment will arrive at (9:30 + 40 minutes) = 10:10 AM .