03 Jun 07
1 edit
Originally posted by abejnoodi am taking it to assume that the first half of the distance has been covered, let's call this total distance D.
Sorry? What do you mean?
This:
A truck travels 15 mph for the first half of the distance of a trip.
How fast must it travel in the second half of the distance in order to average 30 mph for the total trip?
According to your "problem" a truck has travelled a distance of D/2 at a constant or average rate of 15 mph.
So it still has the distance D/2 to travel, that is the remainder.
Since D/2 + D/2 = D .. i am still on solid ground.
Unless of course it has travelled a distance of 0, then the flashy calculus infinite speed approach would work. But as far as i can tell, it wouldn't have travelled distance D/2 at an average speed of 15 mph if it hasn't travelled nothing at all.
03 Jun 07
Originally posted by eldragonflyI really don't understand what you're saying, or your problem with the question. It's fairly straightforward. Think of it this way: speed is distance over time. Now to double your average speed, either half your time--impossible-- or double your distance without losing any time. So the truck has to travel at an infinite speed to go the distance without losing any time.
i am taking it to assume that the first half of the distance has been covered, let's call this total distance D.
According to your "problem" a truck has travelled a distance of D/2 at a constant or average rate of 15 mph.
So it still has the distance D/2 to travel, that is the remainder.
Since D/2 + D/2 = D .. i am still on solid ground.
Unle ...[text shortened]... travelled distance D/2 at an average speed of 15 mph if it hasn't travelled nothing at all.
03 Jun 07
Originally posted by agrysonAh, but you spend time going 45 mph during the second time. Try again.
the truck problem...
45mph?
first half average = 15
second half average = 45
add the two averages and divide by two to get the total average speed...
15 + 45 = 60
60/2 = 30
(The average of the average speeds for any journey should be the total average speed)
03 Jun 07
Originally posted by abejnoodAre you insane abejnood? How can you go a certain distance at an average rate, in this case your chosen 15 mph., without
Ah, but you spend time going 45 mph during the second time. Try again.
1) moving at some velocity
and
2) moving a certain distance.
This is a simple algebra problem.
03 Jun 07
Originally posted by agrysonExactly.
the truck problem...
45mph?
first half average = 15
second half average = 45
add the two averages and divide by two to get the total average speed...
15 + 45 = 60
60/2 = 30
(The average of the average speeds for any journey should be the total average speed)
03 Jun 07
Originally posted by abejnoodWrong. 🙄 You have already covered a distance, which you state is half the distance. Your truck still has to cover the remaining distance, which is the other half of the total distance. 😵
I really don't understand what you're saying, or your problem with the question. It's fairly straightforward. Think of it this way: speed is distance over time. Now to double your average speed, either half your time--impossible-- or double your distance without losing any time. So the truck has to travel at an infinite speed to go the distance without losing any time.
04 Jun 07
Originally posted by eldragonflyWhich is why you have to go at infinite speed! We agree that speed or velocity = displacement divided by time. So you have to double your velocity, right?
Wrong. 🙄 You have already covered a distance, which you state is half the distance. Your truck still has to cover the remaining distance, which is the other half of the total distance. 😵
Assume the distance is 2 miles. You go one mile at 15 mph. So it takes four minutes. Now, in order for the average speed for the trip to be 30 miles an hour, the truck has to travel one mile every to minutes, right? Well, four minutes have already passed, and the truck is only one mile far. So the truck needs to go one mile, and has no time to get there. That's why it needs to travel infinitely fast: it has to go a mile with no time passing at all.
04 Jun 07
Originally posted by eldragonflyNot so simple, apparently. By the way, looking into the past of this problem some more, Einstein himself originally missed this problem when a friend sent it to him.
Are you insane abejnood? How can you go a certain distance at an average rate, in this case your chosen 15 mph., without
1) moving at some velocity
and
2) moving a certain distance.
This is a simple algebra problem.
04 Jun 07
Originally posted by agrysonNot correct. The point of the question is to illustrate a little mathematical pitfall.
the truck problem...
45mph?
first half average = 15
second half average = 45
add the two averages and divide by two to get the total average speed...
15 + 45 = 60
60/2 = 30
(The average of the average speeds for any journey should be the total average speed)
Let's say the distance is 90 miles (this makes the arithmetic easy, although you will find the same outcome whatever distance you choose - try it).
The question says you travelled the first half - 45 miles - at 15mph, so that took you 3 hours.
If you travel the second half - again 45mph - at 45mph, then that will take you another hour. That's 4 hours in total.
90 miles in four hours is an average speed of only 22.5 mph, not 30mph, so your answer doesn't work.
For an average speed of 30mph, you need to do 90 miles in 3 hours. As you spent your 3 hours on the first 45mph, you have to do the remaining 45 miles in zero time. Hence my "infinite speed" answer.
04 Jun 07
1 edit
Originally posted by eldragonflyLet's say the distance for the whole trip is 30 miles.
Are you insane abejnood? How can you go a certain distance at an average rate, in this case your chosen 15 mph., without
1) moving at some velocity
and
2) moving a certain distance.
This is a simple algebra problem.
Going 15mph for 5 miles takes 5/15 = 1/3 hour
Going 45mph for the remaining 5 miles takes 5/45 = 1/9 hour
Total trip time = 4/9 hour
Total distance traveled = 10 mi
Total rate of speed = 10/(4/9) = 90/4 mph = 22.5mph
Oops...you didn't make it to 30mph! What happened?
Edit: Someone beat me to it!
04 Jun 07
Stop getting all 'specific example' otherwise you'll keep confusing the question.
The question is... Total average x = 30
I have two averages (y, z), ___equally weighted___, one being unknown, the other being 15.
y = 15
z = ?
(y+z)/2 = x
(15 + z)/2 = 30
=> 15 + z = 2.30
=> z = 2.30 - 15
z = 60 - 15 = 45
This will work for any unit chosen, miles per hour or pounds of butter per minute, you name it. It's also a generalised solution.