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anonymous

  • 5 years ago

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

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  1. jagatuba
    • 5 years ago
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    Since Jim is 200m behind John (0.5*400m) then covering that distance in 50s is going to be 200m/50s.

  2. anonymous
    • 5 years ago
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    Thanks for your reply. Actually I should have said angular acceleration, not acceleration.

  3. jagatuba
    • 5 years ago
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    Sorry was in another window.

  4. jagatuba
    • 5 years ago
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    Hmmm. This seems like a physics question.

  5. jagatuba
    • 5 years ago
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    So have you seen an equation that goes something like: \[a=v ^{2}\div r\]

  6. jagatuba
    • 5 years ago
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    or \[a=w ^{2}r\]

  7. anonymous
    • 5 years ago
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    Yes I have

  8. jagatuba
    • 5 years ago
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    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.

  9. anonymous
    • 5 years ago
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    that is correct

  10. jagatuba
    • 5 years ago
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    excuse me velocity not accelleration, but I think you knew what I meant.

  11. jagatuba
    • 5 years ago
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    So you have another equation that goes something like this: \[a= 4\pi ^{2}r \div T ^{2}\]

  12. jagatuba
    • 5 years ago
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    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.

  13. jagatuba
    • 5 years ago
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    We also know the radius of the track 400m.

  14. jagatuba
    • 5 years ago
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    So \[a=\left( 64m/s \right)^{2}\div 400m\]

  15. jagatuba
    • 5 years ago
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    So the angular acceleration should be \[9.30909091m/s ^{2}\]

  16. jagatuba
    • 5 years ago
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    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.

  17. anonymous
    • 5 years ago
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    Thank You anyway. I'll try with the Physics forum

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