The walkway is losing velocity at the rate of 2 fps, signified by the 2 in g(t) = 4 - 2t.
I’ll say that the maximum negative speed of the walkway is 1,500,000 fps, and the maximum speed of the person is (200 times the golden ratio) fps.
Enidigm
1742
Arg! You want me to use math to solve a math problem. I forgot initial velocity.
Someone needs to chagne the title of this thread. This has nothing to do with riddles, damned by god or not.
It’s the only damned math puzzle thread I know of, by God!
RichVR
1745
You did it. You really did it. You bastards! You made into… a math thread. You bastards! Damn you. Damn you all to hell!
These kind of problems are virtually impossible for students unless they know calculus or know “the trick.” I swear problem solving is just learning how to apply a library of tricks. It’s not as bad as that godforsaken two trains and a fly problem at least.
The best part about the leading anecdote is that I honestly cannot tell if Neumann summed the series or not; if anyone could do it, it was him. Reading through his list of discoveries and publications is just “jesus chris, that was him?” repeated over and over. A summary.
Jason, are you saying that the problem IS impossible unless they know calculus or a trick?
gmonkey
1748
Are math teachers in the US forced to use stupid units in their problems?
I tried using furlongs per fortnight, but it didn’t fly.
I’m not Jason, but his solution only used algebra, so calculus isn’t necessary. I would say his reconceptualization of the problem might be pretty tough & non-intuitive for algebra students, & might qualify as a trick.
As an aside, my high school physics teacher managed to get the quantitative relations between distance, velocity, and acceleration across using algebra & graph paper – lots & lots of graph paper. That helped me visualize the continuous decrease in velocity, which looked like a smooth triangle when graphed, rather than a series of steps which would represent discrete changes (& lead to incorrect total distances).
I remember graphing lots of stuff in my high school classes of analytic geometry & pre-calculus, but I don’t remember it in algebra I or II. If you don’t have your students practice graphing equations, I think it might be near impossible without a re-orienting trick like Jason used.
I do have them practice it, but I look forward to explaining to them that they can’t operate as though the velocity changes abruptly every second, and then seeing if any of them can pull a Jason sort of move!
Enidigm
1752
That’s why forgetting about time 0 hung me. I was working it out as a table rather distractedly but would have caught myself if I had seen the problem of measuring the distance from time 0 to time 1.
Even worse was that I tried to integrate 6-t and realized I had forgotten how to do it. Ouch.
Here’s a riddle; if the lottery is a tax on people who are bad at math, and Enidgm is bad at math as shown above, how much money did he spend buying Powerball tickets at $2 a ticket?
robsam
1754
I just feel awful reading the comments savaging Dave Perkins and his lack of math teaching skills. I have long considered him my Mortal Enemy, but it pains me to see others disparage him here.
On the other hand, he openly admits that he teaches math. Somehow over all of these years of lurking and sniping at him, I never knew that. Through the magic of the internet, searching for a suitable synonym for Nemesis, a word I refuse to use for him as he is fond of that word, I was presented with Bugbear. I searched no further, and did not look up a proper definition. It fit.
I hereby consider Dave Perkins to be my Bugbear for his crimes against humanity by playing multiple nerd games on the internet and forcing red blooded American children to learn things that they “will never use in real life”, as we used to say. Except for the ones that do. That have good jobs.
I forget why we are mortal bugbears!
Yeah, more or less. Unless they can re-frame it like indicated, I can’t think of a way off the top of my head to figure it out unless they invent just enough of integration or approximation on the fly to get it. I’m an analytics guy though, maybe there’s some immediately obvious way it pops out graphed.
robsam
1757
Of course you don’t remember Dave. That’s exactly the response I expected from you. Just know that I am always watching you, in a non-creepy stalkerish way, and one day you will slip up and I shall give you the most vicious internet zing the world has ever seen!
I leave you all now to your silly math problem, like math has ever done anything good in the world. Fools!
If you graph
f(t) + g(t) = 6 - t
and determine that the area under the line represents distance traveled (because distance = speed * time), then you can find the answer by finding the area of the triangle created by the graphed line.
Area = (1/2) base * height = 18 feet.
As for the time it takes to reach the 18 foot mark, the x-intercept at x = 6 gives the answer, so it takes double that (i.e. 12 seconds) for the person to return to the walkway’s entrance.
For me, this seems like the hardest part of the problem to grasp if you don’t know calculus. The way your problem was set up, the time the man takes from entrance to stop point (velocity = 0) is equal to the time it takes him to go from the stop point back to the entrance. Students might correctly solve the problem without calculus because they mistakenly believe that moving objects always behave that way; for example, a ball thrown into the air takes the same amount of time to get to its maximum height as it does to return to the thrower. Thus, if you want to compute total travel time, just double the time of one leg of the journey.
But that’s not always true of moving objects. The only reason it works in the given problem (& in the case of the ball under gravity) is the man is experiencing constant acceleration of -1 ft/s^2 (acceleration is the derivative of his velocity 6-t, for those keeping track at home). Any problem using non-constant acceleration would result in different times for each leg of his journey.
Ex: Change the problem so the man’s velocity is f(t) = 2 + t, but the walkway’s velocity is g(t) = 4 - t^2. Then the equations change to:
combined velocity = f(t)+g(t) = 6 + t - t^2, meaning
total distance = 6t + .5t^2 - (1/3)*t^3, and
acceleration = 1 - 2t [no longer constant]
The answers in this case are:
- Time needed for man to stop moving forward (velocity = 0) = 3 seconds
- Total time needed for man to move to maximum distance & return (distance = 0) = 5.06 seconds
- Return time from max distance back to entrance = 5.06 - 3 = 2.06 seconds
So under non-constant acceleration, the time from start to max distance won’t be equal to the return time. I’ve never taught algebra, but I suspect that some students may inadvertently come up with the right answer for this problem, but be using faulty reasoning (e.g., a decelerating object that moves away from you will always have equal times for both the forward & return legs).
P.S. Thank you for providing me with a fun diversion from grading stacks of papers. Solving a math problem is so satisfying because it’s so darn certain, rather than all this messy social science stuff.
Enidigm
1760
Sidd’s variant reminded from of one from 7th grade that so infuriated me that i still remember it today, at least in broad outline.
A commercial airliner is travelling at maxiumum speed 550 mph east from its point of departure in a 50mph headwind 1,000 miles to its destination, and then flying back again in the same conditions. Disregarding take-off and landing times, how long will it take the plane to make a round-trip?
…
Got it yet? Well, you’re wrong. The FAA won’t allow airliners to exceed their maximum speed.
