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C4 Math Problem

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Old Jan 16, 2005 | 05:04 PM
  #21  
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Originally Posted by Bilyk
wtf?
the lincoln is doing 1mile/45 sec
which is 1.333333... miles / minute
which is 80 miles / hour

why would it be 90?!?!

1.333333miles/min X 60min/hr = 79.99998 miles/hr

You are wrong sir! No 'rounding off' on this forum! no way...
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Old Jan 16, 2005 | 05:22 PM
  #22  
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Originally Posted by VFR750
Actually, he can't do it. If it's a 1 mile track, it took him 2 minutes to go around it at 30 MPH. To average 60 for both laps he would have to take 2 minutes total. He's out of time! Sorry, kind of a trick question.
Suppose one instead would want average 45 mph for both laps? What would the speed for the second lap be?
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Old Jan 16, 2005 | 05:25 PM
  #23  
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Originally Posted by vetteweiser
I was thinking it would be 90. So why is it 80?
Because you covered the mile in 3/4 of a minute - so essentially you went 1/3 of a mile for every 15 seconds

So if you drove at that speed for 1 minute you would cover 1 1/3 miles per minute

and 1.3333 miles /minute X 60 minutes / hour = 79.99 miles /hour(rounding error)
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Old Jan 16, 2005 | 06:16 PM
  #24  
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Originally Posted by VFR750
Actually, he can't do it. If it's a 1 mile track, it took him 2 minutes to go around it at 30 MPH. To average 60 for both laps he would have to take 2 minutes total. He's out of time! Sorry, kind of a trick question.
That would have worked, if you'd stated it was a 1 mile track. But you didn't.
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Old Jan 16, 2005 | 06:44 PM
  #25  
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Originally Posted by steve9899
That would have worked, if you'd stated it was a 1 mile track. But you didn't.
I doesn't matter how long the track is. I used a mile as an example.
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Old Jan 16, 2005 | 06:47 PM
  #26  
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Originally Posted by steve9899
That would have worked, if you'd stated it was a 1 mile track. But you didn't.

This result is actually independent of track length. The equations for his problem are (correct me if I'm wrong...)

60t = 30t1 + V*t2 (1)
60t = 60t1 (2)
t = t1 + t2 (3)

(2) -> t = t1, put this result in (3) and t2=0... assuming track length > 0 then t2 has to be greater than zero. In other words, no matter how fast you second lap is the total average speed will be less than 60 mph.
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Old Jan 16, 2005 | 06:49 PM
  #27  
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Originally Posted by Nathan Plemons
This was in the newspaper this morning and a little bit of an *** kicker. The problem is easy to solve after you think about it but the obvious answer that you want to say at first is incorrect.

If your ricer is traveling 60 MPH it takes 1 minute to travel 1 mile.
If your Corvette is traveling 120 MPH it takes 30 seconds to travel 1 mile.

How fast must your Lincoln be going to travel the mile in exactly 45 seconds?
Lincoln LS V8 or V6?
Lincoln Town Car or Navigator?
How far can a Lincoln travel at 80 mph in 45 seconds?
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Old Jan 16, 2005 | 07:01 PM
  #28  
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Originally Posted by Blower91
Suppose one instead would want average 45 mph for both laps? What would the speed for the second lap be?
I come up with 90.
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Old Jan 16, 2005 | 07:46 PM
  #29  
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Originally Posted by VFR750
I come up with 90.
So did I! 90 is a little bit higher than I would have guessed before doing the calculation.
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Old Jan 16, 2005 | 08:18 PM
  #30  
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I hate these kids and their "new math".
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Old Jan 16, 2005 | 08:19 PM
  #31  
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Wow. all of this makes me wish i had spent more math classes IN the school and not out Behind the school
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Old Jan 16, 2005 | 09:47 PM
  #32  
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Uh, for you guys who say it's 79.99999... You are the ones introducing the rounding error. Leave it as a fraction and it works perfectly.

.33333 is simply a representation of 1/3. By truncating it at all you are introducing the error. Fractions are always more accurate.
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Old Jan 16, 2005 | 10:22 PM
  #33  
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Quiz #2:

A Corvette driver goes around a city block. On the first side, he goes 10mph... turns, goes 20mph. Goes 30 mph on the third side, then 40 mph, when he reaches his starting point.

What is his average speed?

19.2 mph

Last edited by CentralCoaster; Jan 16, 2005 at 10:27 PM.
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Old Jan 16, 2005 | 11:20 PM
  #34  
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Originally Posted by CentralCoaster
Quiz #2:

A Corvette driver goes around a city block. On the first side, he goes 10mph... turns, goes 20mph. Goes 30 mph on the third side, then 40 mph, when he reaches his starting point.

What is his average speed?

19.2 mph
Yes, 19.2 MPH, assuming he loses no speed in the turns - a fair assumption!
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Old Jan 16, 2005 | 11:24 PM
  #35  
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damm i am going to print this and post it for my algebra classes

these kids today need to see why you need math

i will have to tell them they can't drive fast cars unless they know what a 45 sec between mile posts translates too
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Old Jan 16, 2005 | 11:46 PM
  #36  
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Originally Posted by deiaggie


alot of you guys are trying way too hard on this one...

there are 3600 seconds on an hour (60*60) all you need to do is divide 3600 by the amount of seconds you want to do the mile in and you get how many miles you will do in one hour or MPH

3600/45 seconds=80 MPH
3600/60 seconds=60 MPH
3600/30 seconds=120 MPH
3600/40 seconds=90 MPH
Man I do this all the time, except that I work from the other way...I use my stopwatch to time my miles and then figure my speed (I don't trust my speedometers in my cars)

3600/51 seconds = 70.5 MPH
3600/48 seconds = 75 MPH

It works and I have yet to get a ticket when I stay within the 5 mph buffer that the KHP (unofficially) gives you.
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Old Jan 16, 2005 | 11:53 PM
  #37  
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Old Jan 16, 2005 | 11:54 PM
  #38  
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this thread =
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Old Jan 17, 2005 | 03:42 AM
  #39  
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Originally Posted by Nathan Plemons
Ok, here's the logic, set it up as a fraction.

60 MPH / 60 Seconds vs 120 MPH / 30 Seconds

Well you think that 45 seconds is half way between 30 and 60, 90 mph is half way between 60 and 120, must be right huh?

Unfortunately that's not the case.

To go from 60 to 120 you double it. To go from 60 to 30 you take half of it. Hmm, see the pattern? Anything you do to the top you have to take the INVERSE of on the bottom and vice versa. So 45 seconds is 3/4 of 60 seconds. This means you have to multipy the top by 4/3. 60*(4/3) = 80

Annyoying huh? My dad and I both said 90 first too. He's a math major from way back. He figured it out in another way, which basically accomplishes the same thing.
Too dumb for opera, too smart for nascar!!
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Old Jan 17, 2005 | 12:37 PM
  #40  
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Originally Posted by VFR750
A driver runs 30 MPH for his first lap (in his MG I guess). How fast does he have to go for the second lap to bring his average up to 60 MPH for the two laps? :o
299,792,458 meters per second would do it thanks to relativity (time would stand still for the observer).

Or is that only for a straight line?
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