# Football and Trick Questions

There’s a stereotype that it’s declasse for us intellectual aesthetes to enjoy football, but I don’t care. I enjoy it anyway. Whether you spent any time this weekend watching football or not, I’d like to pose a quick and (maybe!) easy vaguely football-related problem to exercise your brain to make up for all the boob tube time:

You’re standing on the goal line of a standard football field and walk the 100 yards to the opposite goal line at a uniform speed of 2 miles per hour. Upon reaching the other side, you immediately turn around and jog back. How fast do you have to jog so that your average speed over the whole round trip is 4 miles per hour?

Remember: average speed is total distance divided by total time.

Bonus: Rework the problem by developing a nonstandard definition of “average speed” that better fits this particular question.

1. #1 Uncle Al
October 26, 2009

Use perturbational string theory to model 10^(50,000) acceptable scenarios, none of which can be tested.

(300 ft)(60 min/hr)/(5280 ft)(2 mph) = 1.704545… min

(600 ft)/[(1.7 min) +(x min)] = 352 fpm
600 = 352(1.7045455 + x)
352(1.704545…) = 600
x = 0 seconds

Organikers don’t do theory. Organikers do yield.

2. #2 Luke
October 26, 2009

Assuming that the measurement of the field is off by ~ 1nm, you can travel at around the speed of light back. Then, you’ve effectively travelled 200 yards in 1 min 45, (the time it takes to go 100 yards), and attain 4 mph. If we allow an unrealistic scenario of perfect accuracy in measurement, the problem is impossible; we must travel with infinite speed.

3. #3 bdirnbac
October 26, 2009

I’ll post a high school solution so as not perturb the readers (did the Tappet brothers from NPR use this as a problem?).
Traveling the full round trip at 2 mi/hr takes (let’s say) T seconds. Going half that distance takes T/2 seconds.
Traveling the full round trip at 4 mi/hr takes half as long as above, or T/2 seconds (double the speed, halve the time for a given distance). So the slow poke cannot ultimately double his speed from 2 -> 4 mi/hr, because he’s already taken T/2 seconds.

4. #4 RyanR
October 26, 2009

Hmm. Doesn’t look like he can without going faster than light.

I did it this way. First, figure out how long he has to make the 200yd trip if he does it at 4mph. I got 1.7045 minutes.

So T = T1 + T2, where T1 is the trip down and T2 is the trip back. T1 = X1/V1, or 100yd*hr/2mi. With all the unit conversions, I come out with 1.7045 minutes.

So he ate up all the time he had getting down there.

5. #5 Eric Lund
October 26, 2009

@Luke: But if it’s 1 nm off in the wrong direction, then you have already taken more time to walk across the field than you are allowed to take for the round trip. Nonetheless, you provide a clue: if we measure time by the elapsed time in the runner’s frame, returning at the speed of light actually does the trick since no time would elapse for the runner on the return trip. (We also have to assume things like the ability to turn and accelerate to light speed on a dime, but that’s an engineering problem, not a physics problem.)

The first time I saw this problem, it was in a completely different guise, commuting by automobile in traffic. Do it long enough, and sooner or later something like this will happen to you.

6. #6 CRM-114
October 26, 2009

Seeing that you would have zero time to make the return trip, I would advise delegating this to someone who doesn’t mind failures.

October 26, 2009

Since we’re talking football…

You need a friend to kick the football to you just as you reach the goal line on the first leg, so you can start the return trip on the 20-yard line. At this point the problem has a non-infinite, non-relativistic solution.

Rt

8. #8 Tom
October 26, 2009

Hmmm. The blog post is dated Oct 29, and yet my calendar and the responses above say Oct 26. Methinks Matt has an interesting solution to the problem.

9. #9 Andre
October 26, 2009

I’m not sure if this is what you meant by your bonus, but I like the answer better if you average the speed relative to distance traveled. I then get ((2 mph*100 yd)+(x mph*100 yd))/(200 yd) = 4 mph

Solve for x gives you 6 mph for a distance averaged speed.

***Disclaimer: Sorry if this doesn’t make sense or has blatant math errors. This work was done by a synthetic chemist.***

10. #10 Gray Gaffer
October 26, 2009

Well, I worked it from

n/2 + n/x = 2n/2

which reduces to

n/x = 0

so even exceeding the speed of light won’t do it.

Verified the result by noting that the time at 4 mph was already consumed going half the distance at 2 mph.

11. #11 Tophe
October 26, 2009

Your average speed would need to be 6 mph.

12. #12 magista
October 26, 2009

I hit my students with problems like this regularly to drive home the average speed/velocity formula.

But can I just say I pity those who actually have to tackle this in non-metric units? Way to make it more complicated than necessary…

13. #13 Matt Springer
October 26, 2009

Weird about the date thing, but I think I’ve fixed it.

As for the units, I generally use metric around here but in this case the units are immaterial, since it’s a math rather than physics problem.

As for saying “speed”, that’s deliberate and in fact I think #11 has it backward. The average velocity of any round trip is zero, because the total displacement is zero. The average speed isn’t, since it’s the integrated distance rather than displacement in the denominator.

In fact my preferred way to answer the bonus problem is to average the speed with respect to distance rather than time. In that case the required speed is 3 MPH, which is sort of the “intuitive” expectation.

14. #14 Andre
October 26, 2009

Um did you mean 6 mph?

15. #15 Mark Eichenlaub
October 26, 2009

You lose the important properties of the average speed if you average with respect to distance.

If I make two journeys, one going up a 100m field at 2m/s and down at 6m/s, and the other going both ways at 4m/s, they both then have the same “average speed”, but the first takes 67 seconds and the second takes 50 seconds, so the average speed cannot be used to compute the time of the journey given its length, or the length of the journey given its time.

Similarly, consider going up at 10/3 m/s and back at 5m/s. That journey takes 50 seconds – same as the constant 4m/s journey. But the average speed of this new journey would be 25/6 = 4+1/6 m/s. So two different journeys that start and end at the same place and cover the same route in the same amount of time have different average speeds.

16. #16 Mark Eichenlaub
October 26, 2009

To be even more extreme, consider two people walking right next to each other. The follow precisely the same route and walk shoulder-to-shoulder the entire time, except for one millimeter of the route. For that one millimeter, the first person covers it at an extremely fast pace (faster than light if necessary, thinking of this as a classical problem). Then by doing that, we can give that walker any average speed we want – even thousands of miles per hour, just by having him cover one tiny portion of the route extremely quickly. But even though the second walker’s average speed can be made completely arbitrary, that walker follows almost exactly the same path through (classical) spacetime as the first walker, and never gets more than one millimeter of deviation at a given time.

17. #17 dWj
October 26, 2009

If I go 15 MPH, my average speed will round off to 4 MPH. As you say, though, it’s math, not chemistry.

18. #18 Paul Murray
October 26, 2009

Let “s” be “hours per mile”. To get hours per mile over two distances

t1 = s1 d1 — time = hours per mile times distance
t = s1 d1 + s2 d2
d = d1 + d2
so
s = (s1 d1 + s2 d2) / (d1+d2)

solve for s2
s2 = (s(d1+d2) – s1 d1) / d2

Plug in the values. Remember that we have to take reciprocals of mph. Never mind the scaling factors, for now.

s2 = ( 1/40 (1+1) – 1/20 1 ) / 1
= 0

So you have to travel 0 hours per mile.

19. #19 Bob Hawkins
October 27, 2009

Get the ref to throw a flag as you reach the goal line, and give you an untimed play on the return.

20. #20 Gray Gaffer
October 27, 2009

@10: typo fix: n/2 + n/x = 2n/4. sorry bout that.

21. #21 Gray Gaffer
October 27, 2009

to those who think the answer is 6 mph – you’ve already used up the time budget traveling half the distance at half the speed. There is zero time left for the return journey for the average speed to be 4 mph.

rephrasing: n/2 + n/6 != 2n/4.

Note that the problem characteristics are independent of the actual distance travelled. There and back again can not be done at an average of 4 mph, with or without the One Ring.

22. #22 J.J.E.
October 27, 2009

Without peeking, here’s how I formulated it:

Basically, you want to make the time / distance be half (on average) for the whole trip that it is on the first half of the trip. If d is one trip and 2d is the whole trip (some fractional number in units of miles, but since there will be canceling, I won’t bother to figure it out), then you look at it like this (expressing it in units of time):

0.5 ( h / mi ) * d ( mi ) + x ( h / mi ) * d ( mi ) = 0.25 ( h / mi ) * 2d ( mi )

Solving for x, you have to make the second trip in 0 time. So, relativistically impossible.

Two impressions. First, this seems like one of those trick question that relies on the human mind’s poor ability to deal with geometric averages. Secondly, did you interview with Google or Microsoft lately? Sounds like a question they’d ask. (Or possibly McKinsey?)

O.K. Now to paste this into the form and peek at the other responses…

23. #23 nanoAl
October 28, 2009

The round trip is impossible relativistically, but an important amplitude that has to be taken into consideration quantum mechanically. Score one for the inconsistent universe!

24. #24 Gray Gaffer
October 28, 2009

So if n is small enough you might tunnel back?

25. #25 Andy
October 29, 2009

From the perspective of a stationary observer at one side of the football field, then you can’t do. But with relativity, it is possible to do from your perspective (*with one caveat I’ll mention at the end).

Basically, when you travel at any speed, the distance you’re traveling contracts. Thus it takes slightly less time to reach the other side of the field than classical mechanics would say. Then in that *slightly less* amount of time, you have to travel back. Fortunately, if you travel really really close to the speed of light, it takes arbitrarily little time to get anywhere – you’re traveling at roughly c, but the distance you need to go gets smaller and smaller as you approach c.

Here’s my attempt at the math:

When you travel at 2mph, the football field’s length contracts. Let x = sqrt(1-(2mph)^2 / c^2) ~= 1 – 10^-14. Then the length of the field is 100x yards, and it takes you 100x yards / (2 mph) to reach the other side of the field.

That means you have 200 yards / (4 mph) – 100x yards / (2 mph) = (100 yards / 2mph)*(1-x) time to get back. Let y be the speed you take. Then:

y = distance/time = (100 yards * sqrt(1- y^2/c^2)) / ((100 yards / 2mph)*(1-x))

y = 2mph * sqrt(1- y^2/c^2) / (1-sqrt(1-(2mph)^2/c^2))

Approximating:

y^2 = 4 mph^2 *(1 – y^2/c^2) / (10^-28)

y^2 * (1+ (4mph^2/c^2)*10^28) = 4mph^2 *10^28

y^2 = 4mph^2 *10^28 / (1+ (4mph^2/c^2)*10^28)

And y is going to be roughly c – 2mph / 10^14. (I *think* I got the magnitude right on that… don’t have a calculator nearby.)

*Caveat 1: It’s been awhile since I studied relativity, so my math may be off. The principle of length contraction still applies though.

*Caveat 2: I have no idea how general relativity changes the answer. I assumed that the acceleration at the halfway point didn’t happen.