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DenverChief 02-27-2009 09:31 AM

Math question
 
Solve this problem and end the debate please....rep or Casino Cash or both for whoever gets the right answer first


you have 2 vehicles, vehicle A and vehicle B. Vehicle A is 10 miles ahead of vehicle B and traveling at 50 MPH. Vehicle B is traveling at 60 MPH in the same direction as vehicle A, how long does it take for vehicle B to catch vehicle A?

MagicHef 02-27-2009 09:32 AM

1 hour.

Mr. Kotter 02-27-2009 09:32 AM

One hour? :hmmm:

Is this a trick question....?

MagicHef 02-27-2009 09:33 AM

You can set your frame of reference as Vehicle A, so Vehicle B is 10 miles behind, traveling towards it at (60-50)=10 MPH.

Mr. Kotter 02-27-2009 09:33 AM

Quote:

Originally Posted by MagicHef (Post 5532894)
1 hour.


Magic....that Ipod touch "card counting" app kicks ass, don't it? ;)

Demonpenz 02-27-2009 09:34 AM

depends if the dixie lee and the duke boys can make the jump

MagicHef 02-27-2009 09:35 AM

Quote:

Originally Posted by Mr. Kotter (Post 5532905)
Magic....that Ipod touch "card counting" app kicks ass, don't it? ;)

I wish I had an Ipod touch.

Mr. Kotter 02-27-2009 09:35 AM

Quote:

Originally Posted by MagicHef (Post 5532911)
I wish I had an Ipod touch.

So what card counting method are you using, then? :shrug:

soundmind 02-27-2009 09:37 AM

What kind of wind resistance are we working with here?

DenverChief 02-27-2009 09:38 AM

well thanks you math whizzes

there had been a debate...another person was arguing that it would ony take 10 mins since vehicle B is traveling at 60 MPH or a mile a minute and vehicle A is traveling at a slower speed

MagicHef 02-27-2009 09:38 AM

Quote:

Originally Posted by Mr. Kotter (Post 5532914)
So what card counting method are you using, then? :shrug:

Wow... your name has absolutely no anagrams. Amazing.

Mr. Kotter 02-27-2009 09:42 AM

Quote:

Originally Posted by MagicHef (Post 5532929)
Wow... your name has absolutely no anagrams. Amazing.

Yeah. Yours has 12....with some interesting variations. Heh.

Coogs 02-27-2009 09:45 AM

Quote:

Originally Posted by DenverChief (Post 5532927)
well thanks you math whizzes

there had been a debate...another person was arguing that it would ony take 10 mins since vehicle B is traveling at 60 MPH or a mile a minute and vehicle A is traveling at a slower speed

This would have only been true if the speed of vehicle B was 0 MPH.

DenverChief 02-27-2009 09:56 AM

Quote:

Originally Posted by Coogs (Post 5532948)
This would have only been true if the speed of vehicle B was 0 MPH.

yeah I didn't know the formula and of course couldn't find anything on the net

MagicHef 02-27-2009 09:59 AM

Quote:

Originally Posted by Mr. Kotter (Post 5532942)
Yeah. Yours has 12....with some interesting variations. Heh.

Yes. Yes indeed.

Dayze 02-27-2009 10:06 AM

Quote:

Originally Posted by DenverChief (Post 5532927)
well thanks you math whizzes

there had been a debate...another person was arguing that it would ony take 10 mins since vehicle B is traveling at 60 MPH or a mile a minute and vehicle A is traveling at a slower speed


I think the 1 hr answer is correct;

the other argument of 10 minutes, is a bit off. They are failing to realize the ground being made up by car B is only .17 mile per minute.

60/60 = 1 mile per minute
50/60 = .83 mile per minute.

1 - .83 = .17.
.17 x 60 minutes = 10.2 miles.


maybe I'm wrong though.

Coogs 02-27-2009 10:53 AM

Quote:

Originally Posted by Dayze (Post 5533022)
I think the 1 hr answer is correct;

the other argument of 10 minutes, is a bit off. They are failing to realize the ground being made up by car B is only .17 mile per minute.

60/60 = 1 mile per minute
50/60 = .83 mile per minute.

1 - .83 = .17.
.17 x 60 minutes = 10.2 miles.


maybe I'm wrong though.

Rounding error. 50/60 = 0.83333333....

Therefore, 1 - 0.8333333..... = 0.166666666...

Rain Man 02-27-2009 10:54 AM

One hour, minus the effects of relativity.

You know that Car B will pass Car A at some point and will never again be caught. You also know that in one hour they will be at the same point, so therefore...

MagicHef 02-27-2009 11:25 AM

Quote:

Originally Posted by Rain Man (Post 5533200)
One hour, minus the effects of relativity.

You know that Car B will pass Car A at some point and will never again be caught. You also know that in one hour they will be at the same point, so therefore...

Hmm... good point. I would say somewhere between 1 hour and 59 minutes, 59.9999999999999 seconds, depending on if you are standing nearby watching or in one of the cars.


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