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.
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
(t-1)(t-3)(t-6) = 0
so 1 to 3
then 3 to 6
then 6 to 7
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
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
so displacement is zero
and distance is
5.33 + 15.75 + 10.42 = 31.5
check my arithmetic !!!!
To find the displacement of the particle, we need to integrate the velocity function over the given time interval.
The displacement of the particle is given by the definite integral of the velocity function from the initial time to the final time.
Let's integrate the velocity function v(t) = t^3 - 10t^2 + 27t - 18 with respect to t.
To find the displacement, we need to evaluate the integral of v(t) over the given interval [1, 7]:
Displacement = ∫[1, 7] (t^3 - 10t^2 + 27t - 18) dt.
Evaluating this integral, we get:
Displacement = ∫[1, 7] (t^3 - 10t^2 + 27t - 18) dt
= [(1/4)t^4 - (10/3)t^3 + (27/2)t^2 - 18t]∣[1, 7]
= [(1/4)(7^4) - (10/3)(7^3) + (27/2)(7^2) - 18(7)] - [(1/4)(1^4) - (10/3)(1^3) + (27/2)(1^2) - 18(1)]
Simplifying this expression, we can calculate the displacement of the particle.