56. Paddling Time in Still Water A canoeist paddles across a river of uniform width.• When he paddles against the current, the trip takes 4 hours.• When he paddles with the current over the same distance, the trip takes 3 hours.Assuming his paddling speed relative to the water is constant, how long would the same trip take if the water were perfectly still (no current)? Added 5 June 2012 Show Hint Show Solution Hint: Let b be his speed in still water and c the speed of the current. Write two equations for the time taken in each direction, then solve for b. Solution: Let the one-way distance be d km, the paddling speed in still water be b km/h, and the current speed be c km/h.Against the current: d / (b - c) = 4.With the current: d / (b + c) = 3.Equating the expressions for d:4(b - c) = 3(b + c)4b - 4c = 3b + 3cb = 7c.Substitute back to find the distance:d = 4(b - c) = 4(7c - c) = 24c.Time in still water:t = d / b = 24c / 7c = 24 / 7 hours ≈ 3 h 26 min.Therefore, the trip would take 24⁄7 hours (about 3 hours 26 minutes) on still water.
Comments (2)
Assuming the distances are all equal and they cancel each other out
Rower+River=3 hours
Rower-River=4 hours
Basic algebra
2xRower=7
Rower alone takes 3.5 hours
The correct equations (rower and river representing respective speeds) would be:
rower+river=distance/3
rower-river=distance/4
2*rower=7*distance/12
time=distance/rower=24/7=~3.43 hours
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