On an oval track, a car averages $30$ M.P.H. (miles per hour) for the first lap (a lap is $1$ mile around the track). At what average speed must that same car drive the 2nd lap on that same track so that the average speed of both laps is $60$ M.P.H.?
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1This is a classical wrong solution – Alexander Vigodner Nov 17 '14 at 16:56
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1@JohnD: This properly belongs on physics.SE. Average speeds don't add up like that. – MSalters Nov 17 '14 at 16:56
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A wrong solution is funny but a "classical" wrong solution is even funnier. – David Nov 17 '14 at 17:17
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The time available to complete the second lap is zero. 60 MPH is the limit average speed for both laps as the average speed for the second approaches $\infty$. – Thumbnail Nov 17 '14 at 17:23
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This seems like a good problem to give high school math students. I think the average person would get it wrong. I suspect the most likely answer would be $90$. – David Nov 17 '14 at 18:32
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Average speed = Distance / Time = harmonic mean of two speeds
60 = (2*30*x)/(30+x); 30+x=x; 0=30
This is not true, so there is no solution.
I'm not sure if this is correct, so if you want to check my work...
Pakquebchsoflwty
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I'll give Pakquebchsoflwty the checkmark cuz it appears he/she needs it more than Mario G but Mario G's answer and post are actually better. Anybody object to that? – David Nov 17 '14 at 17:20
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First lap lasts $1/30$ hour, let $v_2$ M.P.H. the speed for the second lap, a time $t_2=\frac{1}{v_2}$ hours is required, then the average speed of the two laps is $$\overline{v}=\frac{\overbrace{1}^\text{Distance covered in the first lap}+\overbrace{1}^\text{Distance covered in the second lap}}{\underbrace{\frac{1}{30}}_\text{Time needed for the first lap}+\underbrace{\frac{1}{v_2}}_\text{Time needed for the second lap}}=\frac{60v_2}{v_2+30}$$ And $\overline{v}=60$ M.P.H. iff \begin{align*} &&\frac{60v_2}{v_2+30}&=60\\ \iff &&60v_2&=60v_2+1800\\ \iff &&0&=1800 \end{align*} Therefore, it's not possible to attain an average speed of $60$ M.P.H.
David
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Ángel Mario Gallegos
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A friend asked me this question (as a brain teaser I suppose), and I got it right the first time (although my thinking came across some wrong solutions first which I quickly fixed before emailing him my correct answer). Think of it like this. Imagine $2$ cars on this oval track, call them cars A and B. A drives the first lap (mile) at $30$ MPH average speed, therefore taking $2$ minutes to drive the first $1$ mile. Since the question is asking for an average speed of $60$ MPH over $2$ laps ($2$ miles), we know that it takes $2$ minutes to drive $2$ miles at $60$ MPH. Therefore there is no speed (not even the speed of light) for the 2nd lap of car A that will give it an average speed of $60$ MPH for the $2$ miles because after $2$ minutes, car A is only at the $1$ mile marker and has $1$ more mile to go!
It may help if you imagine car B going a constant $60$ MPH, after $2$ minutes car B would be done the "race" but car A would only be halfway, so even if it had nitrous, a supercharger, dual turbochargers... there is no way it will ever get the $60$ MPH average speed for the $2$ miles with that slow start.
Moral of the story... don't be a "pussyfooter" when it comes to racing cars.
Now had I changed the problem to $40$ MPH for the first lap, then there WOULD be a solution.
David
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$2$ comments: $1$: It is amazing how many "regular" (not math oriented) people get this wrong, giving answers like $60$ or $90$. $2$: There is a "solution" if we "misinterpret" average to mean strictly the average of the $2$ average speeds, disregarding the duration at those speeds. For example, $2$ minutes at 30 MPH and $40$ seconds at $90$ MPH will cover $2$ miles in $2$ min $40$ seconds which is less than $60$ MPH average speed by the normal definition, but if we don't care about time taken, then ($30$+$90$)/$2$ does indeed equal $60$. – David Nov 17 '14 at 17:16