a little math problem

[lucasorion]

here's I came up with it:

I thought in terms of time=t

I used 50 miles an hour as the 1st leg rate, so total time for the trip is 1 hour + whatever time the second leg must be (t)

total distance/total time = rate

100/(1+t) = 2x

2x + 2xt = 100

using 50 miles an hour as x, 100 + 100t = 100

solves to t = 0 (must take 0 time for second leg)

 

[akubi]

Originally posted by: ArchCenturion
<garbage>

Hope this helps

nice try, but you fail math. thanks for playing.

 

[jman19]

Originally posted by: Syringer

Originally posted by: Inspector Jihad

Originally posted by: mugs

Originally posted by: Inspector Jihad

Originally posted by: mugs
Wait, no that's not right

OK, I'm confident it's not possible. You'd have to complete the second half of the trip instantaneously.

Suppose you complete the first half of the trip in 1 hour at 50 mph. You'd then have to adjust your speed to finish with an average speed of 100 mph. You have to travel a total distance of 100 miles in a total time of 1 hour, and you've already used your whole hour.

lol, are you serious

Post your answer.

read the question wrong.

lol, are you serious

Inspector Jihad = owned

 

[ArchCenturion]

Originally posted by: akubi

Originally posted by: ArchCenturion
<garbage>

Hope this helps

nice try, but you fail math. thanks for playing.

Heh i read your post, and it makes sense. I was just came up with something else that seems to work. Would you mind pointing out where I failed?

 

[ArchCenturion]

Could anyone point out at which point i failed?

 

[lucasorion]

you can't choose values for t and the rate (x), t is dependent on the rate.

look at my equation - you can't take longer than 0 to make that second leg, or your rate for the entire trip will be less than 2x. If you make the first leg in one hour, 50mph, and the second leg in 3 minutes, going 1000 mph, your total time will be 63 minutes, at a rate of about 95 mph (1.9x). If you do the first leg in 5 hours(10 mph) and the second leg in 1 minute(3000 mph) your rate for the trip is about 1.99x

 

[akubi]

Originally posted by: ArchCenturion

Originally posted by: akubi

Originally posted by: ArchCenturion
<garbage>

Hope this helps

nice try, but you fail math. thanks for playing.

Heh i read your post, and it makes sense. I was just came up with something else that seems to work. Would you mind pointing out where I failed?

once you picked t to be 1, your x has to be 50. plug those in and see where it fails

 

[thesurge]

If you don't want to use harmonic mean (for style points),

R=D/T

2X=100/(50/X+50/Y)

2X=(2X)(Y/(X+Y)) <---- OMG?

 

[archiloco]

funny how i solved it, i used a drawing to physically draw the problem, came up with instantaneous or impossible. glad i solved it correctly.

 

[Golgatha]

Originally posted by: Mike2002
I'd have to say 3X as well.

 

[ArchCenturion]

Originally posted by: lucasorion
you can't choose values for t and the rate (x), t is dependent on the rate.

look at my equation - you can't take longer than 0 to make that second leg, or your rate for the entire trip will be less than 2x. If you make the first leg in one hour, 50mph, and the second leg in 3 minutes, going 1000 mph, your total time will be 63 minutes, at a rate of about 95 mph (1.9x). If you do the first leg in 5 hours(10 mph) and the second leg in 1 minute(3000 mph) your rate for the trip is about 1.99x

OH Ok that makes sense now, guess I fail, good thing im done with all my math classes.

 

[hellokeith]

Originally posted by: Random Variable
The distance from point A to point B is 100 miles. If you average X mph for the first fifty miles, what speed would you have to average for the second 50 miles so that your average speed for the entire trip is 2X mph?

2x average from the perspective of the traveler or a third party observer?

Theoretically speaking, from the perspective of the traveler, if he/she traveled at the speed of light for the second half of the trip, he/she would experience no passage of time, and thus would average 2x for the trip.

Theoretically speaking, from the perspective of a third party observer, the traveler could enter a worm hole or Nth dimension that would allow him/her instantaneous colocation at the half-way point and the end point, thus averaging 2x for the trip.