You drive to work along a two lane road (one lane each direction) with no passing, stop lights or stop signs. One morning, you encounter no other cars traveling in the same direction as you, either in front of you or behind you. How does your average speed most likely compare with the average speed of the average speeds of the all the other cars on that same road at the same time traveling the same direction (the ones that you didn't encounter)?
Originally posted by ark13Average speed of all cars is most likely the same. Cars ahead are either moving at the same speed or accelerating; cars behind are either moving at the same speed or falling behind. In the latter case, speeds should average out.
You drive to work along a two lane road (one lane each direction) with no passing, stop lights or stop signs. One morning, you encounter no other cars traveling in the same direction as you, either in front of you or behind you. How does your average speed most likely compare with the average speed of the average speeds of the all the other cars on that same road at the same time traveling the same direction (the ones that you didn't encounter)?
Originally posted by HandyAndyThat's the intuitive answer, but I don't believe that's correct. Think about the no-passing restriction.
Average speed of all cars is most likely the same. Cars ahead are either moving at the same speed or accelerating; cars behind are either moving at the same speed or falling behind. In the latter case, speeds should average out.
Well, by no-passing and no-encounters, then i suppose you mean that you are going faster on average than the next car behind you that happens to be slowing all the others down, while you simply may not be catching up to the next car in front because you happen to be going the same speed. however, i don't see where there is enough information in this puzzle to come up with a real solution. i'll be interested to learn what it is if it exists.
I think what the question might be getting at is that cars don't all necessarily travel at the same speed. Therefore a no-passing restriction reduces the average speed, as the faster cars are slowed down.
If you encounter no vehicles, then you are unaffected by the restriction. So your speed is probably higher than average (assuming you have an average speed car to start with).
But it could just be that the question isn't very clear 🙂
Given the limited information the answer must be that traffic behind is travelling at your average speed (or less) an that traffic in front is travelling at your average speed (or more).
To be pedantic one cannot say ANYTHING about average speed of other traffic unless we have an initial start position and know how long scenario will exist for (forever?)
I failed Fluid Dynamics but I think the solution to the (correctly set) problem is in that branch of Math.
I made this up myself, and I just realized I phrased it badly. What I meant to say was "how does your average speed compare to the speed that the cars would go if they were alone on the road. The answer I was looking for is that you're most likely to be going slower than the average cars. Since multiple cars are likely to be limited to a lower speed by a single car they are following then you're probably going lower than the intended speed of the other cars.
Terrible puzzle, I'm sorry. It made sense in my head.
Originally posted by ark13Well, at least it's a good way to avoid getting speeding tickets.
I made this up myself, and I just realized I phrased it badly. What I meant to say was "how does your average speed compare to the speed that the cars would go if they were alone on the road. The answer I was looking for is that you're most likely to be going slower than the average cars. Since multiple cars are likely to be limited to a lower speed by a si ...[text shortened]... ntended speed of the other cars.
Terrible puzzle, I'm sorry. It made sense in my head.
Originally posted by ark13no apology needed. i think multiple answers went directly to that point, including mine.
I made this up myself, and I just realized I phrased it badly. What I meant to say was "how does your average speed compare to the speed that the cars would go if they were alone on the road. The answer I was looking for is that you're most likely to be going slower than the average cars. Since multiple cars are likely to be limited to a lower speed by a si ...[text shortened]... ntended speed of the other cars.
Terrible puzzle, I'm sorry. It made sense in my head.
So, here's one along the same lines: If a cat chases a dog around a bush, what kind of a cat is it likely to be? in fact, it helps to think about what kind of a dog is being chased and why.
Originally posted by coquetteThe cat is a 500-pound Bengal tiger and the dog is a six-pound Chihuahua. Obviously, the cat has a fondness for Mexican food.
no apology needed. i think multiple answers went directly to that point, including mine.
So, here's one along the same lines: If a cat chases a dog around a bush, what kind of a cat is it likely to be? in fact, it helps to think about what kind of a dog is being chased and why.
...if you put the cheese on the bread first and then the ham is it not a cheese and ham sandwhich ?...so why is it always called a ham and cheese sandwhich ?...and does it make a difference if you walk around a bush while
thinking about this problem ?...and will the dog be following you to get the
cheese and the cat to get the ham ?...and does either of the other think of
it as a cheese and ham or ham and cheese sandwhich ?....there are no multiple answers to this problem.
Originally posted by reinfeldham is made of pigglies
...if you put the cheese on the bread first and then the ham is it not a cheese and ham sandwhich ?...so why is it always called a ham and cheese sandwhich ?...and does it make a difference if you walk around a bush while
thinking about this problem ?...and will the dog be following you to get the
cheese and the cat to get the ham ?...and does either of t ...[text shortened]... a cheese and ham or ham and cheese sandwhich ?....there are no multiple answers to this problem.