John L. answered 02/18/21
Naval Academy graduate with more than 10 years experience in teaching
Well, it must be one strong current, or the numbers are not copied into this problem correctly. However, it can still be easily solved. Let C = speed of the current (mi/h) and B = speed of Boat (mi/hr). Since the trip to the Bahama's took the shortest time we set the positive direction of the current C towards Bahamas. Therefore:
110 = 2.5(B+C)
Since return trip is same distance, but current is not fighting against boat, we make C negative or
110 = 323(B-C)
Because the trip distances are the same, the equations both = 110 so
2.5(B+C) = 323(B-C)
2.5B + 2.5C = 323B -323C
325.5C = 320.5B
C = 320.5B/325.5
Then take C and sub into either equation. I selected first
110 = 2.5(B+C)
110 = 2.5(B+320.5B/325.5)
110 = 2.5(1.98464B)
110 = 4.96B
B = 22.17 mph
So solve for C, plug B back into either equation. I came up with 21.89 mph for the current.
Cole S.
I’m so sorry about the typo it’s supposed to be 3 and 2/3 not 323 mph. Hopefully that makes sence now02/18/21