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Topic: Racecars (Read 412 times) |
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ThudnBlunder
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Three cars A,B,C race around a circular track at 290, 130, and 70 kph, respectively.
If one lap is 6.9 km, when all three cars are next abreast of each other
1) how many laps will each car have clocked up?
2) how much time will have elapsed?
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| « Last Edit: Jul 4th, 2005, 11:56am by ThudnBlunder » |
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JiNbOtAk
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Re: Racecars
« Reply #1 on: Jul 4th, 2005, 9:02pm » |
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Do they all start from the same point ? And at the same time ?
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ThudnBlunder
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Re: Racecars
« Reply #2 on: Jul 4th, 2005, 9:36pm » |
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Yes. (Sorry, I forgot to mention that.)
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| « Last Edit: Jul 5th, 2005, 1:09am by ThudnBlunder » |
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Ajax
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Re: Racecars
« Reply #3 on: Jul 5th, 2005, 12:23am » |
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A generalized solution would be: (for A, B, C the speeds and L the track length)
We’d have to find such a time t for which
decimal(A/L*t)= decimal(B/L*t)= decimal(C/L*t) or
decimal(A/L*t-B/L*t)=decimal(A/L*t-C/L*t)=decimal(B/L*t-C/L*t)=0 or
decimal((A-B)/L*t)+decimal((A-C)/L*t)+ decimal((B-C)/L*t)=0 or
(A-B)/L*t=integer and (A-C)/L*t=integer
In the present case: (290-130)/6.9*t=integer and (290-70)/6.9*t=integer and (130-70)/6.9*t=integer
Or 160/6.9*t=integer and 220/6.9*t=integer and 60/6.9*t=integer
Obvious solution: t=6.9 hours
However, maximum common divider of 160, 220,60: 20
Better solution t=6.9/20= 0.345 hours
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| « Last Edit: Jul 5th, 2005, 2:10am by Ajax » |
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mmm
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ThudnBlunder
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Re: Racecars
« Reply #4 on: Jul 5th, 2005, 1:29am » |
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on Jul 5th, 2005, 12:23am, Ajax wrote:
Changing what? 
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markr
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Re: Racecars
« Reply #5 on: Jul 5th, 2005, 1:40am » |
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1)
A: 14.5
B: 6.5
C: 3.5
2) 0.345 hours
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Ajax
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Re: Racecars
« Reply #6 on: Jul 5th, 2005, 2:09am » |
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on Jul 5th, 2005, 1:29am, THUDandBLUNDER wrote:
Changing what? |
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Sorry, I had made a mistake and as I was changing it, I had to do something else, by which time you and markr posted.
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raprap
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Re: Racecars
« Reply #7 on: Jul 7th, 2005, 12:18pm » |
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In 6,9 hrs Car A does 290 laps, Car B does 130 laps, and Car C does 70 laps.
In the same 6.9 hrs
Car A laps Car C 220 times
Car A laps Car B 160 times
Car B laps Car C 80 times
220=11*20
160=8*20
80=4*20
6.9hrs/20=0.345 hrs
so in 0.345 hrs
Car A laps Car C 11 times
Car A laps Car B 8 times
Car B laps Car C 4 times
in 0.345 hrs
Car A does 14.5 laps
Car B does 6.5 laps
Car C does 3.5 laps
so every 0.345 hrs Cars A,B, and C find themselves nose to nose
AND
every 0.690 hrs Cars A, B, and C find themselves nose to nose on the starting line.
Rap
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