A submarine let Hawaii two hours before an aircraft carrier. The vessels traveled in opposite directions. The aircraft carrier traveled at 25 mph for nine hours. After this time the vessels were 280 miles apart. Find the submarines speed.

v1 = 25 mph the rate of the aircraft

v2 = the submarine's speed

t1 = 9 hr the time of the travel of the aircraft

t2 = the time of the travel of the submarine 

t2 = t1 + 2

t2 = 9 + 2

t2 = 11 hr

d = 280 mi the distance between the aircraft and the submarine

Since distance = speed*time we have:

v1*t1 + v2*t2 = d

25*9 + 11v2 = 280

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click here to see the equation solved for v2

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v2 = 5 mph

The submarine's speed was 5 mph.