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anonymous

  • one year ago

The velocity function, in feet per second, is given for a particle moving along a straight line. v(t) = t2 − t − 132, 1 ≤ t ≤ 15 Find the total distance that the particle travels over the given interval

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  1. anonymous
    • one year ago
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    Integrate v(t), but when you solve v(t) = 0 and you'll see there's a root at t = 12, so break up the integral. \[\left| \int\limits_{1}^{12}v(t)~dt \right|+\left| \int\limits_{12}^{15}v(t)~dt \right|\]

  2. anonymous
    • one year ago
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    I got 3084.667 why is the not right

  3. anonymous
    • one year ago
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    I don't know what you did. can you put up your work?

  4. anonymous
    • one year ago
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    \[\int\limits \frac{ t^{3} }{ 3 }-\frac{ t^{2} }{ 2 }+132t\] thats what I used for my integral then I plugged in 1 & 12, then subtracted the 1 value from the 12 value. Then I plugged in 15 and subtracted the 12 value from that & then added the two subtractions

  5. anonymous
    • one year ago
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    \[\left| \frac{ t^3 }{ 3 }-\frac{ t^2 }{ 2 }-132t \right|_1^{12}=\left| -1080-(\frac{ -793 }{ 6 }) \right|=\frac{ 5687 }{ 6 }\] \[\left| \frac{ t^3 }{ 3 }-\frac{ t^2 }{ 2 }-132t \right|_{12}^{15}=\left| -\frac{ 1935 }{ 2 }-(-1080) \right|=\frac{ 225 }{ 2 }\]

  6. anonymous
    • one year ago
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    add them together to get 381/3 ≈ 1060.3 ft

  7. anonymous
    • one year ago
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    but 12 is the zero so you would be adding that integral

  8. anonymous
    • one year ago
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    yes, you want to add the two area to get the total distance traveled|dw:1443626627072:dw|

  9. anonymous
    • one year ago
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    so that isn't the right answer, it's also telling me that it isn't right haha

  10. anonymous
    • one year ago
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    try it without the absolute value signs. -5687/6 + 225/2

  11. anonymous
    • one year ago
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    but then that would be the dispacement

  12. anonymous
    • one year ago
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    i literally got the same thing.. 1060.33 i don't know why it isn't right

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