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anonymous

  • 4 years ago

A train slows down at a constant rate as it rounds a sharp circular horizontal turn. Its initial speed is not known. It takes 18.1 s to slow down from 80 km/h to 33 km/h. The radius of the curve is 163 m. As the train goes around the turn, what is the magnitude of the tangential component of the acceleration? Answer in units of m/s2

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  1. anonymous
    • 4 years ago
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    So its represented by a change in distance over time.... therefore 80-(33km/hr)/18.1 s thn convert because its asking for m/s^2

  2. anonymous
    • 4 years ago
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    (13.06m/s)/18.1s = .721 ms^-2

  3. anonymous
    • 4 years ago
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    Then As the train goes around the turn, what is the sign of the tangential component of the acceleration? Take the direction in which the train is moving to be the positive direction. I just use the sign of the previous answer which is positive? right?

  4. TuringTest
    • 4 years ago
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    1) this is not a question of distance 2) you can't just divide by time again to get m/s^2, you need to follow the physics all acceleration is in m/s^2 in SI units the formula you need here seems to be\[a_{tan}=\frac{\Delta v}{\Delta t}\]

  5. TuringTest
    • 4 years ago
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    ...not sure why the radius of the curve is given

  6. TuringTest
    • 4 years ago
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    note: you will need to convert km/hr into m/s

  7. anonymous
    • 4 years ago
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    isnt the velocity 80km/hr - 33km/hr ????

  8. anonymous
    • 4 years ago
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    47 km/hr converts to 13.06 m/s

  9. TuringTest
    • 4 years ago
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    that is delta v, but you said that was distance in your answer and yes to your conversion, so I guess you had the right answer, you just phrased it strangely for me looks good now though :D

  10. anonymous
    • 4 years ago
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    I just spoke incorrectly... thankyou

  11. anonymous
    • 4 years ago
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    what about the second part of the question

  12. anonymous
    • 4 years ago
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    is it positive?

  13. anonymous
    • 4 years ago
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    It will be negative because it slows the train.

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