A particle moves according to a law of motion s = f(t) = t3 - 12t2 + 21t, t 0, where t is measured in seconds and s in feet.
v(t) = 3t^2 - 24t 21
a(t) = 6t - 24
e) Find the total distance traveled during the first 8 s.
i) When is the particle speeding up? (Enter your answers in ascending order. If you need to use -∞ or ∞, enter -INFINITY or INFINITY.)
( , ) U ( ,infinity)
When is it slowing down?
(0, ) U ( , )
e) Substitute t = 8s into
S = t^3 -12t^2 + 21t
i) Speeding up = acceleration > 0
when is a > 0?
It's when 6t - 24 > 0
Slowing down = acceleration < 0
(just as extra info: constant speed would be when your acceleration = 0)
To find the total distance traveled during the first 8 seconds, we need to calculate the distance traveled between each change in velocity (acceleration).
Let's first find the times when the particle changes its velocity:
To find the velocity, we can differentiate the position function with respect to time.
v(t) = 3t^2 - 24t + 21
Now, let's find when the particle changes its velocity by setting v(t) = 0 and solving for t:
3t^2 - 24t + 21 = 0
We can factor this quadratic equation:
(t - 3)(3t - 7) = 0
Setting each factor equal to zero:
t - 3 = 0 => t = 3
3t - 7 = 0 => t = 7/3
So, the times when the velocity changes are t = 3 and t = 7/3.
Now, we need to find the distance traveled during each interval: (0, 3), (3, 7/3), and (7/3, 8).
To find the distance between two points (t1, t2), we can calculate the definite integral of the velocity function v(t) within this interval, which gives us the change in distance: ∫(t1 to t2) v(t) dt.
Let's find the distance traveled during each interval:
1. (0, 3):
∫(0 to 3) v(t) dt = ∫(0 to 3) (3t^2 - 24t + 21) dt
Integrating each term:
= t^3 - 12t^2 + 21t | (0 to 3)
= (3^3 - 12(3)^2 + 21(3)) - (0^3 - 12(0)^2 + 21(0))
= 27 - 108 + 63 - 0 + 0 + 0
= -18
2. (3, 7/3):
∫(3 to 7/3) v(t) dt = ∫(3 to 7/3) (3t^2 - 24t + 21) dt
Integrating each term:
= t^3 - 12t^2 + 21t | (3 to 7/3)
= ((7/3)^3 - 12(7/3)^2 + 21(7/3)) - (3^3 - 12(3)^2 + 21(3))
= -62/27
3. (7/3, 8):
∫(7/3 to 8) v(t) dt = ∫(7/3 to 8) (3t^2 - 24t + 21) dt
Integrating each term:
= t^3 - 12t^2 + 21t | (7/3 to 8)
= (8^3 - 12(8)^2 + 21(8)) - ((7/3)^3 - 12(7/3)^2 + 21(7/3))
= 120
Finally, to find the total distance traveled during the first 8 seconds, we add up the distance traveled during each interval:
Total distance = -18 + (-62/27) + 120 = 223 5/27 feet.
Now, let's analyze when the particle is speeding up or slowing down:
To determine when the particle is speeding up or slowing down, we need to calculate the acceleration function from the given position function:
a(t) = 6t - 24
Acceleration is positive when a(t) > 0, indicating speeding up, and negative when a(t) < 0, indicating slowing down.
To find when the particle is speeding up, we set a(t) > 0 and solve for t:
6t - 24 > 0
t > 4
So, the particle is speeding up for t > 4.
To find when the particle is slowing down, we set a(t) < 0 and solve for t:
6t - 24 < 0
t < 4
So, the particle is slowing down for t < 4.
Therefore, the particle is speeding up for t > 4 and slowing down for t < 4.
To find the total distance traveled during the first 8 seconds, we need to calculate the displacement of the particle during that time interval.
Displacement is given by the definite integral of the velocity function v(t) with respect to time from 0 to 8.
First, let's calculate the integral of v(t):
∫(3t^2 - 24t + 21) dt = t^3 - 12t^2 + 21t + C
Now we can determine the displacement of the particle by evaluating the integral at the upper and lower limits:
Displacement = [t^3 - 12t^2 + 21t] from 0 to 8
= (8^3 - 12(8)^2 + 21(8)) - (0^3 - 12(0)^2 + 21(0))
= (512 - 12(64) + 168) - (0 - 0 + 0)
= 512 - 768 + 168
Therefore, the total distance traveled during the first 8 seconds is |512 - 768 + 168| = |-88| = 88 feet.
Next, to determine when the particle is speeding up, we need to find the values of t for which the acceleration function a(t) is positive.
The acceleration function is given by a(t) = 6t - 24.
To find when a(t) > 0, we set the equation equal to zero and solve for t:
6t - 24 > 0
6t > 24
t > 4
So the particle is speeding up when t > 4.
Now let's find when the particle is slowing down. The particle will be slowing down when the acceleration is negative (a(t) < 0).
To find when a(t) < 0, we set the equation equal to zero and solve for t:
6t - 24 < 0
6t < 24
t < 4
So the particle is slowing down when t < 4.
In interval notation, the particle is speeding up on (4, ∞) and it is slowing down on (-∞, 4).