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anonymous

  • 4 years ago

A student drives by a stationary police car at 41.7m/s, exactly 7s later the policeman starts after the speeding car, accelerating at a steady 3.75m/s2. How long after the police car begins moving does it take to catch up with the speeding car. i got 34.8s but its wrong, can anyone explain it. Xcar= 291.9+41.7t Xpolice= 1.875t2 Xcar=X police 1.875t2-41.7t-291.2 t=34.8s

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  1. anonymous
    • 4 years ago
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    equation i used was Xf= Xi +Vit+ (1/2)at^2

  2. anonymous
    • 4 years ago
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    Is 291.9 the distance the car travels in seven seconds?

  3. anonymous
    • 4 years ago
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    yea

  4. anonymous
    • 4 years ago
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    Looks good. Double check that you used the correct solution to the quadratic.

  5. anonymous
    • 4 years ago
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    Maybe its just a typo in the solution manual......it has 23.8s

  6. anonymous
    • 4 years ago
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    The math gods seem to agree. http://www.wolframalpha.com/input/?i=0.5*3.75*%28t-7%29%5E2+%3D+41.7*t

  7. mathmate
    • 4 years ago
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    I get 27.833 s.

  8. anonymous
    • 4 years ago
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    @mathmate did you use the same equation?

  9. mathmate
    • 4 years ago
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    I used your equation (corrected) and got 27.833, which is the same answer as eashmore's answer by subtracting 7 seconds.

  10. mathmate
    • 4 years ago
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    Can you double check the data (speed, acceleration, 7 seconds, etc.)

  11. mathmate
    • 4 years ago
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    Note that the clock starts when the police car moves, which is t=7 in eashmore's equation.

  12. anonymous
    • 4 years ago
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    yea data is correct.......since the police car moves 7 seconds later i have cars position as t+7

  13. anonymous
    • 4 years ago
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    Here is a better representation : http://www.wolframalpha.com/input/?i=0.5*3.75*%28t%29%5E2+%3D+41.7*%28t%2B7%29

  14. anonymous
    • 4 years ago
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    ok, thanks a lot

  15. mathmate
    • 4 years ago
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    Guess the answer key has a typo!

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