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Another "Phisics Question" ... sort of

Humor/Fun (Jokes, riddles, and puzzles) · 13 of 13 known posts recovered

johnbyrne · Dec 09, 2008 06:31 PM

#0 source
Well maybe not physics but math.

You want to qualify for the big Go-Cart race. The track is one mile around and you need to make 2 laps with an average lap speed of 30 mph. You get ready to go and your choke sticks so you make the first lap with an average speed of 15 mph. Just as you pass the start the choke opens and you take off. Your cart has a secret engine that allows you to drive well over 100 mph and in fact close to 200 mph. [photo not recovered: devil.gif] Now you don't want the competition to know this so you only want to go as fast as required to qualify. What average speed should you shoot for on the 2nd lap to qualify?
"Often Wrong" .... wife

LNeumann · Dec 09, 2008 06:58 PM

#1 source
I think warp speed would be the answer. You would need to go where no man has gone before. But even 7 times the speed of light won't cut it. You have already exceeded your time limit for two laps in just completing one.

It may have been best to have waved off the attempt. (We get to do that with our airplanes, you know.)
Leonard Neumann
Indianapolis, Indiana, USA

Jim Thomerson · Dec 10, 2008 12:52 PM

#2 source
Two miles at 30 MPH taks four minutes. One mile at 15 MPH takes four minutes. So you need to make the second mile at light speed.

LNeumann · Dec 10, 2008 02:12 PM

#3 source
>Two miles at 30 MPH taks four minutes. One mile at 15 MPH
>takes four minutes. So you need to make the second mile at
>light speed.

Did you notice the "just as you pass the start" comment? That would put it slightly over the 4 minute mark at that point. Can you imagine the acceleration that would be involved to get up to warp speed the second go around. (And to think, the guys on Star Trek don't even wear seat belts when they take off.)
Leonard Neumann
Indianapolis, Indiana, USA

dennis · Dec 10, 2008 04:23 PM

edited#4 source
>>Two miles at 30 MPH taks four minutes. One mile at 15
>MPH
>>takes four minutes. So you need to make the second mile
>at
>>light speed.
>
>Did you notice the "just as you pass the start"
>comment? That would put it slightly over the 4 minute mark at
>that point. Can you imagine the acceleration that would be
>involved to get up to warp speed the second go around. (And
>to think, the guys on Star Trek don't even wear seat belts
>when they take off.)


There is no time factor mentioned in this problem

buildAndFly · Dec 10, 2008 09:44 PM

#5 source

>
>There is no time factor mentioned in this problem

Well then I'm going to say 45 mph average speed for the 2nd lap.

Jim

LNeumann · Dec 11, 2008 08:39 AM

#6 source

>>Did you notice the "just as you pass the start"
>>comment? That would put it slightly over the 4 minute
>>mark at that point. Can you imagine the acceleration
>>that would be involved to get up to warp speed the
>>second go around. (And to think, the guys on Star Trek
>>don't even wear seat belts when they take off.)
>
>
>There is no time factor mentioned in this problem

Oh, but there is a time factor. And you can't just average it as one lap at 15 mph plus one lap at 45 mph equals two laps at an average of 30 mph. Doesn't work that way.

Look at the problem again:
"You want to qualify for the big Go-Cart race. The track is
one mile around and you need to make 2 laps with an average
lap speed of 30 mph."

You want to go two laps at an average speed of 30 mph. Each lap is one mile long, so at 30 mph average it would take you 2 minutes per lap. But if you did the first lap (plus "just over the finish line") at 15 mph, it would take you four minutes (plus) to make that first lap plus. Since you are only allowed 4 minutes total for two laps to make an average of 30 mph, your time is up. Can't be done (not even with warp speed on Star Trek)
Leonard Neumann
Indianapolis, Indiana, USA

Jim Thomerson · Dec 11, 2008 12:39 PM

#7 source
Distance = time x speed. So time is always part of any problem involving distance and speed.

buildAndFly · Dec 11, 2008 01:26 PM

#8 source
>...so at 30 mph average it would take you 2 minutes per lap.

Nope, at 30 mph CONSTANT speed it would take 2 minutes per lap.

Jim

LNeumann · Dec 11, 2008 05:44 PM

#9 source
>>...so at 30 mph average it would take you 2 minutes per
>lap.
>
>Nope, at 30 mph CONSTANT speed it would take 2 minutes per
>lap.
>
>Jim


Even the Indy guys are never constant. I am thinking of taking a lap and they give you the time and calculate your speed from there. Maybe we need a different term. It is not constant but the guys doing the time trials will tell you "the average speed was..."
Leonard Neumann
Indianapolis, Indiana, USA

dennis · Dec 11, 2008 07:28 PM

#10 source
>>>Two miles at 30 MPH taks four minutes. One mile at
>15
>>MPH
>>>takes four minutes. So you need to make the second
>mile
>>at
>>>light speed.
>>
>>Did you notice the "just as you pass the start"
>>comment? That would put it slightly over the 4 minute
>mark at
>>that point. Can you imagine the acceleration that would
>be
>>involved to get up to warp speed the second go around.
>(And
>>to think, the guys on Star Trek don't even wear seat
>belts
>>when they take off.)
>
>
>There is no time factor mentioned in this problem


Sorry guys but I'm being a wise a!!. In the real world the driver would have pulled off the track, fixed the problem and then tried to requalify. There would be no time factor as there would be no time

buildAndFly · Dec 11, 2008 10:18 PM

#11 source
You guys win, but it bothers me that the formula (lap_1_mph + lap_2_mph)/2 doesn't give the correct answer. I'm not a math wiz, but I assume it has something to do with the fact that mph represents a relationship and not just a number.

Jim

Wayne C · Aug 14, 2009 06:38 PM

#12 source
>You guys win, but it bothers me that the formula (lap_1_mph +
>lap_2_mph)/2 doesn't give the correct answer. I'm not a math
>wiz, but I assume it has something to do with the fact that
>mph represents a relationship and not just a number.
>
>Jim

I realize this post was from a while back but chose to respond anyway.

(speed + speed)/2 isn't the correct formula. You did average two numbers, but the result did describe the situation because it did not take the fixed distance and time into account.

You need (total distance traveled) / (total time used to travel the distance) = distance per time
Wayne C -- Northeast TX
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