# 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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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
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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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