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anonymous

  • one year ago

A particle moves along the x-axis with velocity v(t) = t^2 − 4, with t measured in seconds and v(t) measured in feet per second. Find the total distance travelled by the particle from t = 0 to t = 3 seconds.

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  1. Astrophysics
    • one year ago
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    For the distance you need to know the positive and negative values because if you integrate from 0 to 3 you just have the displacement.

  2. anonymous
    • one year ago
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    so how would you find those

  3. Astrophysics
    • one year ago
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    Well we know v(t) = t^2-4 =(t+2)(t-2), notice this gives us the negative value when v(t) is going in the opposite directions, so you will have to take absolute values of when it goes from \[\left| \int\limits_{0}^{2} t^2-4 dt\right| + \left| \int\limits_{2}^{3} t^2-4dt \right|\] notice that distance is just \[\int\limits_{0}^{3} |v(t)| dt\]

  4. Astrophysics
    • one year ago
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    From 0<t<2 it's negative and 2<t<3 it's positive so we take the absolute value for that reason

  5. anonymous
    • one year ago
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    so the answer would be -3.0333?

  6. Astrophysics
    • one year ago
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    No, remember we are taking the absolute value

  7. Astrophysics
    • one year ago
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    That makes negative = +

  8. anonymous
    • one year ago
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    oh yes so it would be 7.6

  9. Astrophysics
    • one year ago
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    distance can't be negative, it's a scalar value

  10. anonymous
    • one year ago
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    thank you so much!

  11. Astrophysics
    • one year ago
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    Yes, that looks good \[\frac{ 23 }{ 2 } \approx 7.7\]

  12. Astrophysics
    • one year ago
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    So it worked out?

  13. anonymous
    • one year ago
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    yes it did and it made a lot of sense the way you explained it thank you

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