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anonymous

  • one year ago

A particle is moving with velocity v(t) = t2 – 9t + 18 with distance, s measured in meters, left or right of zero, and t measured in seconds, with t between 0 and 8 seconds inclusive. The position at time t = 0 sec is 1 meter right of zero, that is, s(0) = 1. 1.The average velocity over the interval 0 to 8 seconds 2.The instantaneous velocity and speed at time 5 secs Help please

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  1. freckles
    • one year ago
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    you can find the position function by integrating the velocity function and then apply the condition s(0)=1

  2. freckles
    • one year ago
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    you can find the average velocity over the interval t in [0,8] by computing: \[\frac{s(8)-s(0)}{8-0}\]

  3. anonymous
    • one year ago
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    acceleration ->integrate -> velocity velocity -> integrate -> displacement

  4. anonymous
    • one year ago
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    would the average velocity be 3.33

  5. phi
    • one year ago
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    oh wait. first subtract off s(0)= 1

  6. freckles
    • one year ago
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    can you find the instantaneous velocity? (hint: think derivative of position)

  7. freckles
    • one year ago
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    and you are actually already given the derivative of the position :)

  8. anonymous
    • one year ago
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    instantantaneous velocity is dx/dt

  9. anonymous
    • one year ago
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    so you plug in 5 into the v(t)?

  10. freckles
    • one year ago
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    yep

  11. freckles
    • one year ago
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    are you sure @phi that is what I got

  12. anonymous
    • one year ago
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    how do I find the average velocity with initial condition s(0)=1

  13. phi
    • one year ago
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    oh, never mind. that constant is subtracted off. so yes 3.333

  14. anonymous
    • one year ago
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    you can always check your answer on wolframalpha, but phi has a good explanation on it

  15. anonymous
    • one year ago
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    thank you very much!

  16. phi
    • one year ago
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    how do I find the average velocity with initial condition s(0)=1 after integrating v(t) you get \[ \frac{t^3}{3}- \frac{9t^2}{2} +18t + C \] when t is 0 that should be 1, i.e. C= 1 and when we do s(8)- s(0) the one's cancel. so we just need \[ \frac{t^3}{3}- \frac{9t^2}{2} +18t\] evaluated at t= 8 (for the top) the bottom is 8-0= 8

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