A particle moves on a vertical line. Its position, s, in metres at t seconds is given by s(t) = t^3 - 9t^2 + 24t, t>0/
I found the velocity and acceleration functions.
s'(t) = 3t^2 - 18t + 24
s''(t) = 6t-18
b) When is the particle moving up? down?
c) Find the distance the particle travels between t=0 and t=6.
b) To determine when the particle is moving up or down, we need to analyze the velocity function, s'(t) = 3t^2 - 18t + 24.
First, we can find the critical points by setting s'(t) equal to zero and solving for t:
3t^2 - 18t + 24 = 0
Next, we can factor the equation:
3(t^2 - 6t + 8) = 0
Now, we can solve for t by factoring the quadratic equation:
(t - 2)(t - 4) = 0
Therefore, t = 2 or t = 4.
Now, let's analyze the intervals between these critical points.
- When t < 2, the velocity is positive, so the particle is moving up.
- When 2 < t < 4, the velocity is negative, so the particle is moving down.
- When t > 4, the velocity is positive, so the particle is moving up.
Therefore, the particle is moving up from t = 0 to t = 2 and again from t = 4 onwards. The particle is moving down between t = 2 and t = 4.
c) To find the distance traveled by the particle between t = 0 and t = 6, we need to calculate the definite integral of the absolute value of the velocity function from t = 0 to t = 6:
∫[0 to 6] |s'(t)| dt
Since we know the velocity function is piecewise, we need to split the integral into two parts:
∫[0 to 2] |3t^2 - 18t + 24| dt + ∫[2 to 6] |3t^2 - 18t + 24| dt
To solve each integral, we can integrate the function within the absolute value brackets separately:
∫[0 to 2] (3t^2 - 18t + 24) dt + ∫[2 to 6] (-3t^2 + 18t - 24) dt
Evaluating each integral, we get:
[(t^3/3) - 9t^2/2 + 24t] from 0 to 2 + [(-t^3/3) + 9t^2/2 - 24t] from 2 to 6
Plugging in the values, we have:
[(2^3/3) - 9(2^2)/2 + 24(2)] + [(6^3/3) - 9(6^2)/2 + 24(6)] - [(0^3/3) - 9(0^2)/2 + 24(0)]
Simplifying further, we get:
[8/3 - 18/2 + 48] + [216/3 - 9(36)/2 + 144] - 0
= (8/3 - 27 + 48) + (72 - 162 + 144)
= (118/3) + (54)
= 118/3 + 162/3
= 280/3
Therefore, the distance traveled by the particle between t = 0 and t = 6 is 280/3 meters.
To determine when the particle is moving up or down, we need to analyze the velocity function s'(t) = 3t^2 - 18t + 24.
b) When is the particle moving up?
To find the intervals when the particle is moving up, we look for values of t when the velocity function s'(t) is positive (greater than 0). This means the particle is moving upward.
Set s'(t) > 0:
3t^2 - 18t + 24 > 0
To solve this quadratic inequality, we first find the roots by setting the equation equal to 0:
3t^2 - 18t + 24 = 0
Factor the equation:
3(t-2)(t-4) = 0
Now we have two critical points, t = 2 and t = 4.
To determine the intervals when the particle is moving up, we test the intervals on a number line:
Interval 1: (-∞, 2)
Choose a value within this interval, say t = 0
Substitute t = 0 into s'(t):
s'(0) = 3(0)^2 - 18(0) + 24 = 24
Since s'(0) > 0, the particle is moving up in this interval.
Interval 2: (2, 4)
Choose a value within this interval, say t = 3
Substitute t = 3 into s'(t):
s'(3) = 3(3)^2 - 18(3) + 24 = 3
Since s'(3) > 0, the particle is moving up in this interval.
Interval 3: (4, ∞)
Choose a value within this interval, say t = 5
Substitute t = 5 into s'(t):
s'(5) = 3(5)^2 - 18(5) + 24 = -21
Since s'(5) < 0, the particle is moving down in this interval.
Therefore, the particle is moving up on the intervals (0, 2) and (2, 4).
c) Find the distance the particle travels between t = 0 and t = 6.
To find the distance the particle travels, we calculate the total displacement over the given time interval. The distance is always positive, so we need to find the absolute value of the displacement.
The displacement can be found by integrating the velocity function s'(t) over the interval [0, 6]:
∫[0, 6] (3t^2 - 18t + 24) dt
Integrating term by term, we get:
∫[0, 6] (3t^2) dt - ∫[0, 6] (18t) dt + ∫[0, 6] (24) dt
Using the power rule of integration, we can integrate each term separately:
[t^3]^6_0 - [9t^2]^6_0 + [24t]^6_0
Substituting the limits of integration and simplifying:
(6^3 - 0^3) - (9(6^2) - 0) + (24(6) - 0)
Calculating further:
216 - 324 + 144
Now we calculate the absolute value of the total displacement:
|216 - 324 + 144| = |36| = 36
Therefore, the particle travels a distance of 36 meters between t = 0 and t = 6.
Right on the first.
Now when it is going up> When is s'positive? When is it negative? The sign of s'is direction. Hint: factor the quadratic and test domains to see when it is +-
The distance it travels? Integrate velocity over time. The area under a velocity time graph is distance.