Timeline for The motorcyclist's challenge
Current License: CC BY-SA 3.0
6 events
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| Aug 10, 2011 at 6:41 | comment | added | Eric |
and the $<0$ part is too stringent. the exact boundary is $<{a}_{1}$. As long as $<{a}_{1}$ is satisfied, ${a}_{1}$ can always catch up. This inequality defines a subset of B3={(a1,a2,a3)|2a2≤a1+1,a3<1}. So that characterization still holds.
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| Aug 10, 2011 at 6:28 | comment | added | Barry Cipra | @unknown, your comment came in while I was posting my second edit. The real difference between us is that you did the algebra correctly. I finally realized something was amiss in my formula for the average speed when it occurred to me that it implied a negative average speed for $a_2 = 1/2$ even when $a_3$ is close to 1, which is patent nonsense. | |
| Aug 10, 2011 at 6:19 | history | edited | Barry Cipra | CC BY-SA 3.0 |
added 492 characters in body
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| Aug 10, 2011 at 6:19 | comment | added | Eric | @Barry: Your idea is correct. Actually I considered this approach, too. The only difference is that I also took the distance $L$ into my calculation. $L$ cancels out in the end, as one would expect, leaving me with $(3{a}_{2}{a}_{3}+{a}_{2}+{a}_{3}-1)/(3+{a}_{2}+{a}_{3}-{a}_{2}{a}_{3})<{a}_{1}$, which defines a subset of my characterization of ${B}^{3}$. | |
| Aug 10, 2011 at 4:00 | history | edited | Barry Cipra | CC BY-SA 3.0 |
correction
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| Aug 10, 2011 at 3:47 | history | answered | Barry Cipra | CC BY-SA 3.0 |