Why is there a non-physical solution to this problem? (quadratic equation)

[pixel]

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Saw this problem the other day and I have a question about the solution(s):

A river is flowing downstream at a speed of 3 mph. A boat travels up the river 24 miles, turns around and travels down the river back to its starting location. If the total time for the round trip is 6 hours, what is the speed of the boat relative to the river (assumed to be the same for the upstream and downstream trips). So choosing upstream as the positive direction, and using v as the speed (magnitude) of the boat relative to the river:

Time for first leg of trip: 24/(v-3)
Time for second leg of trip: -24/(-v-3) = 24/(v+3)
Therefore, 24/(v-3) + 24/(v+3) = 6

Solving the resultant quadratic equation, we get v = 9 mph and -1 mph. Both of these are numerical solutions to the equation, as can be seen by substitution, but the second one is not physically meaningful: v is defined to be positive and anyhow any v < 3 would mean that the boat could never go upstream.

Why does the -1 solution occur? Is it something about how the problem is formulated?

 

[phinds]

Quadratics are just math and quadratics always give two answers. When the quadratic is a solution to a physical problem, it's possible that only one of the solutions is a possible solution to the physical problem. I solved it slightly differently and got a quadratic that showed that one of the times had to be either 2 hours or 12 hours. Now 12 hours was a perfectly fine solution to the quadratic but obvious was not a possible solution to the physical problem, just as was the case with your solution showing a speed of -1 mph.

If the physical problem is finding a point relative to a curve, just as one example, usually BOTH solutions are valid solutions unless there is some physical constraint which invalidates one of them.

 

[pixel]

I solved it slightly differently and got a quadratic that showed that one of the times had to be either 2 hours or 12 hours.

I'm not sure what you solved. The problem stipulated that the time was 6 hours for the round trip. That was a given.

Quadratic equations can have only one solution, when b²-4ac = 0.

 

[TomHart]

I got 9 and -1 for the velocity solutions, which results in time solutions of t1 = 4, t2 = 2 and t1 = -6, t2 = +12. Both solutions result in a 6-hour round trip.

 

[haruspex]

It can help to work through the equations used, plugging in the nonphysical answer.
Going upstream, witn a relative speed of -1mph, the boat travels downstream at 4mph. To get to its destination, it has to do this for -6hours, i.e. time has to be run backwards. Having reached the upstream wharf 6 hours earlier than it left, it now travels downstream at 3-1=2mph and takes +12h to get back to its start point. Total time 12-6=6h.
To read that another way, if the boat leaves the upstream wharf fighting the current at 1mph it takes 12h to go downstream. If,instead, it works with the current it will take only 6h. The difference is 6h. It just happens that the same equations describe both problems, so both solutions emerge.

 

[phinds]

I'm not sure what you solved. The problem stipulated that the time was 6 hours for the round trip. That was a given.

Quadratic equations can have only one solution, when b²-4ac = 0.

ONE of the times. Read what I wrote.

 

[pixel]

It just happens that the same equations describe both problems, so both solutions emerge.

Interesting. And good that this alternate problem was not the one assigned!

ONE of the times. Read what I wrote.

Yes sir !

I solved it slightly differently and got a quadratic that showed that one of the times had to be either 2 hours or 12 hours.

Maybe you can share your solution with us.

 

[phinds]

Interesting. And good that this alternate problem was not the one assigned!Yes sir !Maybe you can share your solution with us.

My notes are illegible but surely you get 2 hours for the downsteam travel time, no?