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The velocity function, in feet per second, is given for a particle moving along a straight line.

v(t) = t3 − 10t2 + 27t − 18,

1 ≤ t ≤ 7

Find the displacement

Find the total distance that the particle travels over the given interval.

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5 answers
  1. integrate to find x(1) and x(7)
    x(7)-x(1) is the displacement vector

    to find the distance scalar we must know when the particle is moving forward and when back and integrate over those intervals adding absolute values

    v = t^3 - 10 t^2 + 27 t - 18
    integral
    = t^4/4-10t^3/3 +27t^2/2 -18t
    at 7 integral = 205278
    at 1 integral = -7.58
    so displacement = 205285
    =============================
    now where is v positive and where negative?
    need zeros of v

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  2. (t-1)(t-3)(t-6) = 0
    so 1 to 3
    then 3 to 6
    then 6 to 7

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  3. I used this to do the numbers:
    http://www.mathportal.org/calculators/calculus/integral-calculator.php

    t = 1 to t = 3 int = +5.33
    t = 3 to t = 6 int = -15.75
    t = 6 to t = 7 int = +10.42
    so distance = -21.5

    this also says my answer for part 1 should have been
    5.33 -15.75+10.42 = -10

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  4. v = t^3 - 10 t^2 + 27 t - 18
    integral
    = t^4/4-10t^3/3 +27t^2/2 -18t
    at 7 integral = -7.58
    at 1 integral = -7.58
    so displacement = 0

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  5. so displacement is zero
    and distance is
    5.33 + 15.75 + 10.42 = 31.5

    check my arithmetic !!!!

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