Question

A fishing barge leaves from a dock and travels upstream (against the current) for 4 hours until it reaches its destination 12 miles away. On the return trip the barge travels the same distance downstream (with the current) in 2 hours. Find the speed of the barge in still water.

Answer 1

v(b) = 4,5 mil/h speed of the barge in still water

Explanation:

d = v*t barge going upstream 12 miles and 4 hours trip

barge returning back 12 miles and 2 hours trip

let call v(b) barge velocity and

v(w) water velocity

d = 12 (Mil) = 4 (h)* [(v(b) - v(w)]

3 = v(b) - v(w) (1)

d = 12 (mil) = 2 (h) * [ (v(b) + v(w)]

6 = v(b) + v(w) (2)

Equations (1) and (2) is a two system equation. Solving

from equation (1) v(w) = v(b) - 3

By subtitution in equation (2)

6 = v(b) + v(b) - 3

9 = 2v(b)

v(b) = 9/2 ⇒ v(b) = 4,5 mil/h