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Topic: Racetrack Laps (Read 7368 times) |
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c0wt00n
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Racetrack Laps
« on: Jul 22nd, 2013, 10:46pm » |
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I searched and did not find a thread for this puzzle.
"You are at a track day at your local racecourse in your new Porsche. Because it's a crowded day at the track, you are only allowed to do two laps. You haven't driven your car at the track yet, so you took the first lap easy, at 30 miles per hour. But you do want to see what your ridiculous sports car can do. How fast do you have to go on the second lap to end the day with an average speed of 60 miles per hour? "
Am I correct that its not possible? If you want to double your average speed, you basically would have to do both laps in the time it took you to do the first one. Since its not possible to instantaneously do a lap, its not possible to get to a 60mph average. Correct?
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yevvi
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Re: Racetrack Laps
« Reply #2 on: Jun 10th, 2015, 1:29am » |
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I m confused. Why 90mph is not the answer?
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SMQ
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Re: Racetrack Laps
« Reply #3 on: Jun 10th, 2015, 6:37am » |
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Because speeds, being a reciprocal measure (distance over time), don't average that way (unless you keep the time constant).
Let's do the math on your proposed answer, picking the length of the track more-or-less at random to be 1.5 miles (a common length of oval tracks here in 'Murrica; the exact length doesn't matter to the form of the answer):
First lap: 1.5 miles/t1 hours = 30 mph, so it takes you t1 = 0.05 hours = 3 minutes = 180 seconds to do the first lap.
Second lap: 1.5 miles/t2 hours = 90 mph, so it takes you t2 = 0.0166... hours = 1 minute = 60 seconds to do the second lap.
So in total you've driven 3 miles in t1 + t2 = 0.0666... hours = 4 minutes = 240 seconds for an average speed of 45 mph and not the 60 you might expect.
Now let's look at it the other way and see how much time you'd need to take in the second lap to average 60 mph for the whole run (given that t1 = 0.05): 3 miles/(t1 + t2) = 60 mph, so t1 + t2 = 0.05 hours, which means t2 = 0. Uh-oh. So in order to average out at 60 mph, you have to do the second lap instantaneously (or at infinite speed, whichever).
Hope that helps. 
--SMQ
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--SMQ
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Virgilijus
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Re: Racetrack Laps
« Reply #4 on: Jun 10th, 2015, 6:47am » |
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Lets say the track is length D. When he went around the first time, it took him time T1. Knowing that speed is equal to distance travelled divided by time (s = d/t) we get:
(Equation 1) 30 = D / T1.
If we want his average speed for both laps to be equal to 60 mph, we use the same formula. However, the distance traveled will now be twice what it was before.
(Equation 2) 60 = 2D / (T1 + T2)
where T2 is the time for the second lap. We're trying to solve for T2, so let's rearrange Equation 1 to isolate T1 and substitute it into Equation 2.
T1 = D / 30
(Equation 3) 60 = 2D / ( (D/30) + T2)
On the right side, we can multiply the numerator and the denominator both by 30 (really just multiplying by 1) to make the fractions easier.
60 = 60D / (D + 30T2)
Rearrange a little more and we get
D + 30T2 = D
30*T2 = 0
The only way for this to be true is if it takes 0 seconds to complete the second lap which, obviously, is impossible. Therefore, he can't ever double his speed with a second lap.
Also, for some reason I can't use the YABBC tags when posting <:^(
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Grimbal
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Re: Racetrack Laps
« Reply #5 on: Jun 10th, 2015, 2:20pm » |
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The intuition of averaging the 2 speeds can work, but you have to express the speed in hours per mile. (or easier, minutes per mile).
You made the first lap at 2 minutes per mile. Now you want an average of 1 minutes per mile over 2 laps. Obviously you need to do the second at 0 minutes per mile in order to get 1 as the average of 0 and 2.
0 minutes per mile is actually an infinite speed.
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rloginunix
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Re: Racetrack Laps
« Reply #6 on: Jun 10th, 2015, 2:57pm » |
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A proposed tiny problem statement adjustment. In the phrase "so you took the first lap easy, at 30 miles per hour" - what does "30 mph" actually mean? While it is tempting to say "constant speed" the modern view of physics holds that no object with a non-zero mass can change its speed instantly. The Porsche's speed at the start is zero, then it is changing as some function of time. Point is - the problem (imho) should read "... first lap easy, at an average speed of 30 miles per hour ..."
All of the above calculations and suggestions then jive - average the averages carefully.
(And remember that the average velocity in this case is zero )
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yevvi
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Re: Racetrack Laps
« Reply #7 on: Jun 11th, 2015, 1:34am » |
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Oh i see now. Thanks a lot.
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rmsgrey
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Re: Racetrack Laps
« Reply #8 on: Jun 11th, 2015, 4:58am » |
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on Jun 10th, 2015, 2:57pm, rloginunix wrote:A proposed tiny problem statement adjustment. In the phrase "so you took the first lap easy, at 30 miles per hour" - what does "30 mph" actually mean? While it is tempting to say "constant speed" the modern view of physics holds that no object with a non-zero mass can change its speed instantly. The Porsche's speed at the start is zero, then it is changing as some function of time. Point is - the problem (imho) should read "... first lap easy, at an average speed of 30 miles per hour ..."
All of the above calculations and suggestions then jive - average the averages carefully.
(And remember that the average velocity in this case is zero ) |
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Assuming "speed for the first lap" has meaning, that meaning is either "average speed" or "constant speed". For the purposes of the question, it doesn't matter which.
Nothing in the problem statement says you have to start the lap stationary (in fact, since a lap starts when you cross the starting line, starting a lap while stationary is tricky...). If you drive onto the course, drive the two laps, and drive off the course, then your starting speed can be pretty much anything your car can manage...
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