anonymous
  • anonymous
Hi, Could you please help me with the following problem I'm trying to solve without success: John and Jim are racing around a circular track in sports cars. The track has a radius of 400 M. John is leading the race by half a lap and both cars are initially traveling at 60 M/s. Jim than begins to accelerate (acceleration not given). After 50 seconds Jim catches up with John. What was Jim's acceleration ? Thank You
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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jagatuba
  • jagatuba
Since Jim is 200m behind John (0.5*400m) then covering that distance in 50s is going to be 200m/50s.
anonymous
  • anonymous
Thanks for your reply. Actually I should have said angular acceleration, not acceleration.
jagatuba
  • jagatuba
Sorry was in another window.

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jagatuba
  • jagatuba
Hmmm. This seems like a physics question.
jagatuba
  • jagatuba
So have you seen an equation that goes something like: \[a=v ^{2}\div r\]
jagatuba
  • jagatuba
or \[a=w ^{2}r\]
anonymous
  • anonymous
Yes I have
jagatuba
  • jagatuba
Ok so you are dealing with tangential and angular acceleration with those equations right? Those are the two things that are going to make up your angular acceleration.
anonymous
  • anonymous
that is correct
jagatuba
  • jagatuba
excuse me velocity not accelleration, but I think you knew what I meant.
jagatuba
  • jagatuba
So you have another equation that goes something like this: \[a= 4\pi ^{2}r \div T ^{2}\]
jagatuba
  • jagatuba
Since we know that Jim has to increase his velocity by 4m/s to catch up to John in 50 seconds. This would make his tangential velocity 64m/s.
jagatuba
  • jagatuba
We also know the radius of the track 400m.
jagatuba
  • jagatuba
So \[a=\left( 64m/s \right)^{2}\div 400m\]
jagatuba
  • jagatuba
So the angular acceleration should be \[9.30909091m/s ^{2}\]
jagatuba
  • jagatuba
I think . . . It's been a loooong time since physics so take that answer with a grain of salt. Check into the physics forum to be sure I'm getting my equations right.
anonymous
  • anonymous
Thank You anyway. I'll try with the Physics forum

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