A New POV on an Old Problem

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Suppose that you are on a private boat, cruising on a river like you always do. You start at a dock, then steer the boat 1 hour upriver. You then steer the boat back downriver, and go back to the dock. You hear from somebody that your boat's only lifesaver accidentally dropped at the start of your cruise, and has been floating downriver. This causes you to keep on going downriver until you finally catch up to the lifesaver 1 mile downriver from the dock.

The question is: how fast was the river moving?

You may work this out the typical way: setting a variable to be the speed of the boat, setting a variable to be the speed of the river, then solving for it. But there is an easier way.


What is we looked at the point of view of the lifesaver instead o the boat? Relative to the lifesaver, the river is not moving. The boat cruises away for a while, the cruises back to get it. Remember that the river isn't moving, so the boat takes the same amount of time to cruise back to the lifesaver as cruising away. Since it took 1 hour to cruise away, then it takes 1 hour to cruise back, which means that the whole ordeal took 2 hours. The lifesaver floated downriver 1 mile in these two hours, implying the river flows at 0.5 miles per hour.

#Algebra #Speed #Relativity #Trick #PointOfView

Note by Daniel Liu
6 years, 10 months ago

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Comments

My original thought was .5 miles per hour. But then after thinking, I realized we don't know how long the return trip took, which means we don't know how fast the river is moving. We also don't know if the boat returned at the same speed.

Kevin Kacmarcik - 6 years, 10 months ago

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Did you read the solution?

Daniel Liu - 6 years, 10 months ago

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I did. The solution requires a lot of assumptions that are not available in the problem though. It doesn't say how long it takes to get back, and additionally it doesn't say how long the boat takes to go the extra mile. All we know is a boat drove an hour upstream. We can't assume it took the same amount of time to get back, and since we don't know how far the boat traveled in that hour, we can't determine how long it took to go the extra mile past the dock. For the boat to return in the same amount of time, it must have been traveling more slowly, since the river is flowing away from the starting direction. You also say the river is not moving. The river IS moving, it's simply not moving relative to the lifesaver, which is floating along at the same speed. I think it's a great question, and a great solution, I just think the question needs to be worded a bit differently. :-)

Kevin Kacmarcik - 6 years, 10 months ago

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@Kevin Kacmarcik I basically copied the question and solution from "The Art and Craft of Problem Solving" because I thought it was pretty clever.

All we know is a boat drove an hour upstream. We can't assume it took the same amount of time to get back

Perhaps, but we CAN tell that it took the same amount of time to get back to the dock and then go the extra mile.

You can visualize this by instead pretending that the river is solid ground, and the boat is a person. The lifesaver stays stationary on the ground. Meanwhile, the person walks away from the lifesaver. 1 hour later, he starts walking back, until he reaches the lifesaver again.

This is an equivalent scenario to looking at the problem in the point of view of the lifesaver.

If the position of the dock is bothering you (because the position of the dock moves if the river is stationary) you can realize that the dock doesn't even matter at all to the time it takes to get back to the lifesaver; we only need to use the dock to calculate the speed of the river at the end.

Daniel Liu - 6 years, 10 months ago
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