Help transfer an idea/riddle into a formula

[Buckie]

Hi folks. I recently had a long trip, driving and counting the remaining miles and thought up a funny riddle: suppose you drive and reach a sign that says your destination point is 142 kilometers away. You then proceed forward at a speed of 142 kilometers per hour. Next sign comes up and says the destination point is now 141 kilometers away. You immediately drop the speed to 141 kilometers per hour and drive on. You keep going like that and eventually reach a point that says "1 kilometer" and you walk towards the destination at 1 kilometer per hour, and naturally the last 1 km takes 60 minutes. So I wondered if there was a way to put it into a formula or maybe a way to calculate it easily, and some background as to what is may be called. The best I could come up with was arithmetical progression but I'm not sure we're dealing with that here. So... it will be really cool to know. I'm not skilled in mathematics but I like science.

 

Perhaps it may be even more interesting if we make it infinite, i.e. instead of cutting it into 1 kilometer chunks drop the speed gradually, although we'd never reach the destination then.

[Acme]

I think you have described the triangular numbers. 1+2+3+4+... The equation is [math]\frac{n^2+n}{2}[/math]

 

Substituting 142 for n gives 10,153 [hours].

 

PS There is a famous anecdote involving Carl Friedrich Gauss solving the problem of summing an arithmetic progression when he was a child. Here's a list of 109 tellings.>> Versions of the Gauss Schoolroom Anecdote

Edited July 31, 2014 by Acme

[Bignose]

Substituting 142 for n gives 10,153 [hours].

Acme, this fails the "common sense" sniff test. The last km takes 1 hour. In the 2nd to last km you travel faster than in the last km. Ergo the 2nd to last km must take less than 1 full hour. You can use the same logic to know that the 3rd to last km takes less than 1 full hour, the 4th to last km takes less than a full hours, and so on for all 142 km. Since every single km takes 1 hour or less, your final total in time has to be less than 142 hours. 10 thousand is way off.

 

[math]Total time = \sum_{i = 1}^{142} \frac{1}{i}[/math] is how I would calculate it.

Edited July 31, 2014 by Bignose

[Acme]

Acme, this fails the "common sense" sniff test. The last km takes 1 hour. In the 2nd to last km you travel faster than in the last km. Ergo the 2nd to last km must take less than 1 full hour. You can use the same logic to know that the 3rd to last km takes less than 1 full hour, the 4th to last km takes less than a full hours, and so on for all 142 km. Since every single km takes 1 hour or less, your final total in time has to be less than 142 hours. 10 thousand is way off.

 

[math]Total time = \sum_{i = 1}^{142} \frac{1}{i}[/math] is how I would calculate it.

I'll have to mull your response, but the way I read it every interval between signs is driven in 1 hour. At best I may be off 1 interval, i.e. 141 intervals and not 142.

 

Edit: Mulling complete. Bignose is correct insomuch as I mislabeled the units. My answer is for kilometers travelled and not hours taken. In my defense I note the OP does not specify the answer sought. If one thinks of the signs as marks on a ruler, then we start from 142 and cover 142 intervals before we arrive at destination 0. Disregarding the additiona small times it takes to slow 1km/h at each sign, the trip will take 142 hours and cover 10,153 km.

Edited August 1, 2014 by Acme

[Bignose]

the trip will take 142 hours and cover 10,153 km.

Unless I am grossly misreading the OP, the total distance traveled is 142 km. As stated "reach a sign that says your destination point is 142 kilometers away."

 

During the 1 km between the 142 and 141 sign, the rate of speed is 142km/hr. Therefore it takes 1/142 hr to travel that distance.

During the 1 km between the 141 and 140 sign, the rate of speed is 141km/hr. Therefore it takes 1/141 hr to travel that distance.

During the 1 km between the 140 and 139 sign, the rate of speed is 140km/hr. Therefore it takes 1/140 hr to travel that distance.

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During the 1 km between the 2 and 1 sign, the rate of speed is 2km/hr. Therefore it takes 1/2 hr to travel that distance.

During the 1 km between the 1 and 0 sign, the rate of speed is 1km/hr. Therefore it takes 1/1 hr to travel that distance.

 

You add up all of them, as given by the formula I posted above, and you have your answer.

[Acme]

Unless I am grossly misreading the OP, the total distance traveled is 142 km. As stated "reach a sign that says your destination point is 142 kilometers away."

 

During the 1 km between the 142 and 141 sign, the rate of speed is 142km/hr. Therefore it takes 1/142 hr to travel that distance.

During the 1 km between the 141 and 140 sign, the rate of speed is 141km/hr. Therefore it takes 1/141 hr to travel that distance.

During the 1 km between the 140 and 139 sign, the rate of speed is 140km/hr. Therefore it takes 1/140 hr to travel that distance.

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.

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During the 1 km between the 2 and 1 sign, the rate of speed is 2km/hr. Therefore it takes 1/2 hr to travel that distance.

During the 1 km between the 1 and 0 sign, the rate of speed is 1km/hr. Therefore it takes 1/1 hr to travel that distance.

 

You add up all of them, as given by the formula I posted above, and you have your answer.

Acknowledged. Will await Buckie's further edification.

[Buckie]

Yes, it's exactly as Bignose said: the total distance to travel is 142 km and every kilometer there's a sign and at every sign we drop the speed by 1 kilometer per hour, starting at 142 km/hr and ending at 1km/hr.

 

I think this sums it up:

https://www.wolframalpha.com/input/?i=sum&a=*C.sum-_*Calculator.dflt-&f2=1%2Fx&f=Sum.sumfunction%5Cu005f1%2Fx&f3=1&f=Sum.sumlowerlimit%5Cu005f1&f4=142&f=Sum.sumupperlimit2_142&a=*FVarOpt.1-_**-.***Sum.sumvariable---.*--

 

So thank you for educating me on this!

 

I think it's fairly easy to modify the formula to handle different cases, such as speed being reduced only by half (0.5 km/hr instead of 1 km/hr at every sign) - that would transform it into 1/(2*i), correct?

 

 

Another question that I have is can we calculate the time if the speed is being dropped gradually but still according to the same starting principle, starting at 142 km/hr and at 141 km mark already finding ourselves going at 141 km/hr. It's obvious that we're never reaching the destination but is there a way to set a cut-off point somewhere for our measurements? Perhaps at 1km mark (without changing the starting and ending conditions)? And can it also be put into a clear formula?

And a followup question: what is the way to calculate the sum? I don't know how WA do it, but is it possible to have a simple single string like (x+y)/z and have the correct result or maybe the only way to actually produce the number is to calculate intermediate results for each step and then add them up?

[John]

And a followup question: what is the way to calculate the sum? I don't know how WA do it, but is it possible to have a simple single string like (x+y)/z and have the correct result or maybe the only way to actually produce the number is to calculate intermediate results for each step and then add them up?

 

If you're asking about the sum Bignose presented, then no, there isn't a simple formula for calculating the sum. However, we can approximate it fairly well.

 

[math]\sum_{i=1}^{\infty} \frac{1}{i}[/math] is called the harmonic series, and the partial sum [math]\sum_{i=1}^{n} \frac{1}{i}[/math] for some finite n is called the n^th harmonic number, denoted H_n. It turns out that as n approaches infinity, the difference between H_n and ln n approaches a limit, which we call the Euler-Mascheroni constant. Thus, as n increases, the approximation [math]\ln {n} + \gamma[/math] (where [math]\gamma[/math] is the aforementioned constant) becomes more and more accurate.

 

The Calculation section of Wikipedia's article on harmonic numbers gives the first few terms of an asymptotic expansion for H_n, and including these dramatically improves the approximation.

 

For comparison, check WolframAlpha's results for H₁₄₂, ln(142) + gamma, and ln(142) + gamma + 1/(2(142)) - 1/(12(142²)) + 1/(120(142⁴)).

Edited August 1, 2014 by John

[Buckie]

Thank you! I'll have to read more thoroughly on Harmonic numbers and harmonic series.

 

What about my other question about calculating time if the speed is being constantly reduced?

[Bignose]

What about my other question about calculating time if the speed is being constantly reduced?

This can be set up as a calculus problem. "Speed being constantly reduced" translates into an acceleration (deceleration). Then you use the fact that acceleration is the time derivative of velocity and velocity is the time derivative of position.