Thread: Word problem help, linear function (probably!)

The word problem goes like this:

It takes some people 2 hours to row up a river (6,4 km long) and back again (so total distance travelled = 12,8 km). The current is 2,4 km/h. How fast would the boat have travelled if there were no current?

I reason like this: When there is a current, the speed is 6,4 km/h. When the boat travels upstream, the current slows the boat down by 2,4 km/h, and when the boat travels downstream the current speeds up the boat equally much. So the current cancels itself out and the speed is still 6,4 km/h. But that's not the correct answer, the correct answer is 7,2 km/h.

Drawing a picture of the problem and assigning variable names doesn't help me very much, it's just a line with length L = 6,4 units long, a boat with the constant speed x and the current represented by an arrow.

It seems to me that the correct answer is an equation system, one linear equation for the way up and another for the way down, with two variables (time and speed). But that seems a bit too complicated. What am I thinking wrong? Thank you for any help!

Suppose that the speed of the boat without a current is , then the boats speed upstream will be , and downsteam
That means that the times to go upstream and back will be and respectively.
The sum of these we are told is That leads to a quadratic which you should be able to solve. (There turn out to be two answers one of which is

Suppose a physical body travels distance d₁ in time t₁ and then travels distance d₂ in time t₂. The speed on the first segment is v₁ = d₁ / t₁, and the speed on the second segment is v₂ = d₂ / t₂. What can we say about the average speed when t₁ = t₂ = t? It is, as expected, the arithmetical mean of v₁ and v₂, i.e., (v₁ + v₂) / 2. Indeed, the average speed is (d₁ + d₂) / (t₁ + t₂) = (d₁ + d₂) / 2t = (d₁ / t + d₂ / t) / 2 = (v₁ + v₂) / 2.

What can we say about the average speed when d₁ = d₂ = d? Consider an example when d is 1 mile, v₁ = 1 mile per hour and v₂ = .001 miles per hour. Then t₁ = 1 and t₂ = 1000, so the average speed is (1 + 1) / (1 + 1000) < .002, while (v₁ + v₂) / 2 ≈ .5. If v₂ approaches 0, the total time approaches infinity and the true average speed approaches 0, while the arithmetical mean of the two speeds approaches v₁ / 2. Algebraically, the average speed is 2d / (t₁ + t₂) = 2d / (d / v₁ + d / t₂) = 2 / (1 / v₁ + 1 / v₂) = 2v₁v₂ / (v₁ + v₂).

Thank you, this was very helpful! I will use this approach next time I see a similar problem, that is, to think in terms of average value. Naming the time segments was pretty good. I kind of missed out on that the time will be different if no current is involved, so I think next time I see a problem like this I will think about what differs in the two situations. Also I will think about dividing values into segments. Perhaps you can categorize this problem as a with/without problem - modifying a situation so that it fits new circmstances, so that could be a way to recognize the method one should apply. Thank you a lot, again!