Old car speed puzzle

There’s an old puzzle that involves a car on a two-mile track that makes the circuit in four minutes. Or maybe it was a one-mile track and the car took two minutes to make it around. The car needs to make another circuit at such a speed that the total time for both runs would be… something minutes. Most people would assume that the car could double its speed to meet the aggregate time; however the answer is that the car much reach (or exceed) the speed of light in order to do this. The puzzle is a ‘perfect system’ where speeds and times are exact, and speeds are ‘instant’ (i.e., acceleration is not taken into consideration).

As I recall, the puzzle was worded simply enough that I could figure it out if I thought about it, but I’m lazy and I thought someone might know it immediately.

A car makes a lap around a one mile track at an average speed 30 miles per hour - how fast would it need to make the second lap to make the average over both laps 60 miles per hour.

What I remember about that was that you were going after an “average speed” and what you were left with forced you to go at some FTL rate to be able to average whatever was required.

It’s been a long time, but at least I heard of a similar problem. No details, just sympathy.

ETA: Andy L’s answer is closer than mine.

Thanks, guys.

Note that the answer isn’t “the speed of light”, but “infinitely fast; it has to make the entire second lap instantaneously”.

Note also that the puzzle turns over the specific average speed used being the time-weighted average speed (where the speed of the car on each leg is weighted by the proportion of time spent on that leg). The thing is, if you use the distance-weighted average speed (where the speed of the car on each leg is weighted by the proportion of total distance spent on that leg), then doubling the speed for the second leg is the correct answer.

Of course, time-weighted average speed has a very nice property: it turns out to be equal to total distance/total time, as a result of speed itself being a time-weighted measure of distance travelled. Still, it’s not the only notion of “average” one could ever intend to speak of.

It’s important to note, when retelling it to friends and whatnot, that it requires the traveler to already be at the 50% mark. It only works because the traveler needs to think “If I were going twice as fast, I’d be done by now.” There’s no way he can catch his twice-as-fast twin if that twin has already finished the race.

But if the traveler has only gone, say, 49% of the way, then his twin would have only gotten to the 98% mark and there’s still time to catch him. Of course, that doesn’t make it necessarily possible, but it at least gives the traveler a shot.

Another way to look at it, for the mathematically-challenged: at 60 MPH, you travel one mile in exactly one minute. At 30 MPH, you likewise will go one mile in exactly 2 minutes (half the speed means twice the time). But to average 60 for both laps, you need to finish them both in 2 minutes, but you’ve already taken the 2 minutes to do the one-mile lap!

I’ve been waiting for the SAT people to spring this on my students for awhile now, but so far no such luck (even tho each test does have one of these R*T = D problems).

Nitpick: Tripling the speed.

Heh, good point.

This is why I flunked physics.

What are the units of distance-weighted average speed? Not distance per unit time, obviously…

No, it’s still distance per unit time. It’s just measured differently, and isn’t as useful for real-world applications.

Think of it this way: I’m driving along a road, and periodically take a picture of my speedometer (say, once per minute). I then look at all of the readings in those pictures, and average them: This gives me my time-weighted average speed. Well, technically, you’d need to take a pictures at shorter and shorter time intervals, and take the limit as the interval goes to zero, but once a minute is enough for most practical purposes.

Now, instead, suppose that instead of me taking pictures of my dashboard, there are cops stationed along the road (say, one every mile) with radar guns. They all measure my speed as I pass them, and then average all of the readings (again, take the limit as the spacing between cops goes to zero). This is my distance-weighted average speed.

An addendum to that: We usually measure speeds in terms of distance per unit time, but we could (and sometimes do) measure them instead in terms of time per unit distance. For instance, running speeds are often expressed as how many minutes it takes to run a mile. If my speedometer and the cops’ radar guns were set up this way, then it’d be the average of the cops’ measurements that would be relevant to my whole trip, not mine. I imagine that the reason we typically use distance per time instead of time per distance is that stopped things (zero distance in a finite amount of time) are a lot more common than teleporting things (finite amount of distance in zero time), so we avoid having to divide by zero.

But taking average speed isn’t a question of sampling your (instantaneous) speed at various intervals in time - you had me going there for a moment, but it ain’t so. It’s a question of taking your distance run divided by the time; snapshots of your speedo are potentially misleading, but snapshots of your odometer and dashboard clock tell me all I need to know about your average speed. What’s more, a single snapshot at the end of the hour tells me your average speed over the hour with exactly as much accuracy as I would have had from any higher number of pictures over that same hour. True, I could have learned more about your average speeds over the shorter time intervals, if I cared, but I wouldn’t know any more about your average speed over the whole hour.

No, the snapshots of the speedometer actually give you the same answer as the total distance divided by total time. Provided, as I said, that you take your snapshots frequently enough (this avoids problems like stepping on the gas as soon as one picture is taken, and then immediately braking back down just before the next one). In practice, of course, it’s easier to use the total distance and total time, but either method is valid.

As Chronos said, it is indeed distance per unit time. What we’re doing is we’re taking S1 (the speed on leg 1 of the journey) and S2 (the speed on leg 2 of the journey) and applying some weighted average to them to get p * S1 + q * S2. What are p and q? Well, in a time-weighted average, p is the proportion of time on the first leg (time on first leg / total time) and q is the proportion of time on the second leg (time on second leg / total time). In a distance weighted average, p is the proportion of distance on the first leg (distance on first leg / total distance) and similarly for q with the second leg. Either way, p and q are unitless. And S1 and S2 of course have units of distance per time. So overall, p * S1 + q * S2 has units of distance per time, regardless of whether you use a time-weighted or distance-weighted average. They’re just different kinds of averages for different purposes.

Suppose that every piece of the road suffers some amount of wear as you drive over it, the amount of wear it suffers being proportional to the speed at which you drive over it. Then, at the end, you want to know your average speed, as determined by the average wear on the road. In this case, you’d probably be inclined to interpret “average wear on the road” as being the distance-weighted average, and that will correspond precisely to the distance-weighted average speed.

But what if the car is on a treadmill?

Rewording the question with other people’s input:

Two cars begin a two-lap race on an oval track that is 1 mile in diameter. Car A is going at 60 mph, without accounting for acceleration/deceleration. Car B is going at 30 mph. How fast must Car B be traveling in the second lap to beat or tie Car A to the finish line?


Car A finishes the race in 2 minutes. At the end of the first lap, Car B already used 2 minutes to travel the 1 mile at 30 mph. To defeat Car A, Car B must travel fast enough to go backwards in time.

That sounds fairly close to the wording of the problem. Not quite, but then I can’t remember the original wording (hence the question).