Choo-choo
- Where have you been hiding lately? I haven’t seen you for days.
- I have been travelling.
- Really? That explains things.
- I read somewhere that travelling is good for your creativity.
- What will they think of next?! That one thinks with one’s knees?!
- I returned by train. The last leg was 500km long and the train’s average speed was exactly 50km/h.
- How on earth do you know?
- The journey lasted exactly ten hours.
- Well, that explains things.
- I was wondering, did we ever cover a distance along the way 50km long in exactly one hour?
- Of course, you had to. Didn’t you say the average speed was 50km/h?
Problem source: Martin Gardner , December 1979.
June 10th, 2010 at 1:58 am
Let d(t) be the distance covered after travelling for time t. For every moment t of the journey, apart from its first hour, record a(t) = d(t)-d(t-1 hour). If there is a time t for which a(t)=50km, then you covered a distance of 50km in the hour from t-1 hour to t.
a(1 hour) is the distance covered in the first hour. a(2 hours) is the distance covered in the second hour, and so on. Thus, as the whole 500km journey took 10 hours,
a(1 hour) + a(2 hours) + … + a(10 hours) = 500km
There are 10 terms on the left-hand side. If any of them, a(t), is 50km exactly, then you covered a distance of 50km in the hour from t-1 hour to t. If they were each less (resp. greater) than 50km, their sum would be less (resp. greater) than 500km. Thus at least one term is less than 50km and at least one is greater. The function a is continuous (as the train can’t travel a positive distance in zero time). Thus a(t)=50km for some t, after all.
June 10th, 2010 at 2:39 am
This isn’t actually the first method I tried. I recalled a technique used to solve a problem about a man climbing up a mountain one day, then down it using the same path the following day. I tried to adapt the idea to this problem, but I couldn’t see a proof. That’s because what I was attempting to prove isn’t true. The corresponding assertion for a journey of a non-integer number of hours is not true.
The idea is this. Label points along the track by how far they are, in kilometres along the track, from the start of your journey. Let P[0] be the start of your journey and P[500] the end. Suppose you take the A train, and the B train makes a similar journey along the same track except:
* the B train started at P[50]
* the B train started one hour after the A train started
* the B train travelled in such a way that, however long it took the A train to reach P[x] (where x is no more than 450), the B train took the same length of time to cover that same distance from P[50] to P[x+50].
Now the A train has covered the last 50km from P[x-50] to P[x] in 1 hour if and only if the trains are head to head at P[x].
Therefore you covered a distance along the way 50km long in exactly one hour if and only if the trains are ever head to head.
Without exploiting the fact that the total distance travelled, 500km, and the duration of the journey, 10 hours, are multiples of the shorter distance and time mentioned, 50km and 1 hour, I tried to prove that the trains must indeed be head to head at some stage, but failed.
Suppose that the journey had been one of 130km, and 2 hours 36 minutes. Then again the mean speed is 50km/hr. Is it necessarily true that you covered a distance along the way 50km long in exactly one hour? Not if the train travelled as follows:
12:00: Leave P[0]
From 12:00 to 12:48, 10 km at a steady 12.5 km/h
12:48: Reach P[10]
From 12:48 to 13:00, 50 km at a steady 250 km/h
13:00: Reach P[60]. B train leaves P[50], so A train is ahead of B train
From 13:00 to 13:36, 10 km at a steady 16 2/3 km/h
13:36: Reach P[70]. B train won’t even reach P[60] until 13:48, so A train is ahead of B train
From 13:36 to 13:48, 50 km at a steady 250 km/h
13:48: Reach P[120]. B train won’t even reach P[110] until 14:00, so A train is ahead of B train
From 13:36 to 14:36, 10 km at a steady 12.5 km/h
14:36: Reach P[130]. B train is at P[120], so A train is ahead of B train
Thus the A train has kept ahead of the B train, so there is no 1-hour period during which the train covered exactly 50 km.
June 10th, 2010 at 6:15 pm
Simpler approach: let d(t) be the monotonic function describing the distance traveled at time t, so 0 <= d(t) <= 500 for 0 <= t <=10. Certainly the chord connecting [0, d(0)] and [10, d(10)] has slope 50 kph. Then, by the intermediate value theorem, there is an interval [T, T+1] which also has a connecting chord [T, d(T)] and [T+1, d(T+1)] with slope 50 kph.
June 10th, 2010 at 8:38 pm
@Mike
I thought like you at first that it was a simple intermediate value/mean value theorem and then tried like Richard to prove it, till I saw it was true only for time intervals dividing the whole time of the journey (which is the same as saying that the interval is one hour and the journey is an integer number of hours.)
Richard gives a counter-example and it’s quite easy to show there’s always one for non-dividing intervals.
Fact is you can’t use the mean value theorem to prove it, or the intermediate value theorem on other functions like f(t)=d(t)-d(t-1) (which we would like to have a value of 50 somewhere).
Mean value theorem only tells you there’s at least one moment where the speed is exactly 50 kph…