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anonymous

  • 4 years ago

From my previous question : At the moment the train’s speed is 45 km/h, what is the magnitude of the total accelera- tion? Answer in units of m/s2

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  1. anonymous
    • 4 years ago
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    First convert 45 km/hr

  2. anonymous
    • 4 years ago
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    12.5 m/s

  3. anonymous
    • 4 years ago
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    heres where` the r comes from?? v^2/r

  4. anonymous
    • 4 years ago
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    .959

  5. anonymous
    • 4 years ago
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    It doesn't matter what the speed is. The earlier prompt says "slows down at a constant rate" which means the train experiences constant acceleration.

  6. anonymous
    • 4 years ago
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    which is?

  7. anonymous
    • 4 years ago
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    can you explain this further.

  8. anonymous
    • 4 years ago
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    You're right. I overlooked the circular part. \[a = {v^2 \over r}\]

  9. anonymous
    • 4 years ago
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    But that is only the centripetal component, there is still the tangential. We are asked for the total magnitude. \[a = \sqrt{\left(v^2 \over 2 \right)^2 + a_t^2}\]where \(a_t\) is the acceleration value you found in the other question.

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