Sorry this is really long. Just wondering how I would do each of these

A particle is moving with velocity v(t) = t^2 – 9t + 18 with distance, s measured in meters, left or right of zero, and t measured in seconds, with t between 0 and 8 seconds inclusive. The position at time t = 0 sec is 1 meter right of zero, that is, s(0) = 1.

The average velocity over the interval 0 to 8 seconds
The instantaneous velocity and speed at time 5 secs

The time interval(s) when the particle is moving right

The time interval(s) when the particle is
going faster
slowing down

Find the total distance the particle has traveled between 0 and 8 seconds

remember that d(s)/dt = v(t)

and acceleration = v ' (t)

s(t) = (1/3)t^3 - (9/2)t^2 + 18t + c
given s(0) = 1
1 = 0-0+0+c
c = 1

s(t) = (1/3)t^3 - (9/2)t^2 + 18t + c

s(0) = 1
s(8) = 512/3 - 288 + 144+1
= 83/3

avg velocity for first 8 seconds
= (83/3 - 1)/(8-0)
= 10/3 m/s
v(5) = 25 - 45 + 18 = -2 m/s

the particle is moving to the right if v > 0
t^2 – 9t + 18>0
(t-6)(t-3) > 0
So v > 0 for 3 < t < 6

a(t) = 2t - 9

the object is moving faster when a > 0
and slower when a < 0

take over

To solve the given problems step by step, let's start with the first one:

1. The average velocity over the interval 0 to 8 seconds:
The average velocity is defined as the total displacement divided by the total time. In this case, the displacement is the difference in position, and the total time is 8 seconds.

To find the average velocity, we need to calculate the displacement between the starting and ending positions:

s(8) - s(0) = (8^2 - 9*8 + 18) - (0^2 - 9*0 + 18) = (64 - 72) - (0 - 0 + 18) = -8 - 18 = -26

Now we can calculate the average velocity:

Average velocity = (-26 m) / (8 s) = -3.25 m/s

Therefore, the average velocity over the interval 0 to 8 seconds is -3.25 m/s.

2. The instantaneous velocity and speed at time 5 seconds:
The instantaneous velocity at a specific time can be found by taking the derivative of the position function with respect to time (t). In this case, the position function is given by s(t) = t^2 - 9t + 18.

To find the instantaneous velocity at t = 5 seconds, we differentiate the position function with respect to t:

v(t) = d(s(t))/dt = d(t^2 - 9t + 18)/dt = 2t - 9

Now we substitute t = 5 into the derivative:

v(5) = 2(5) - 9 = 10 - 9 = 1 m/s

Therefore, the instantaneous velocity at t = 5 seconds is 1 m/s.

To find the speed at t = 5 seconds, we take the absolute value of the instantaneous velocity:

Speed = |v(5)| = |1| = 1 m/s

Therefore, the speed at t = 5 seconds is 1 m/s.

3. The time interval(s) when the particle is moving right:
The particle is moving right when the velocity is positive. We can find the time intervals by setting the velocity function, v(t), greater than zero:

t^2 - 9t + 18 > 0

To solve this inequality, we can factorize the quadratic expression:

(t - 3)(t - 6) > 0

The sign of the inequality changes when t = 3 or t = 6. We can use these values to create intervals:

(0, 3) U (6, 8)

Therefore, the particle is moving right in the time intervals (0, 3) and (6, 8).

4. The time interval(s) when the particle is going faster:
The particle is going faster when the absolute value of the velocity is increasing. The velocity function is v(t) = t^2 - 9t + 18.

To find the time intervals when the particle is going faster, we need to look for when the velocity function is increasing. This occurs when the derivative of the velocity function is positive:

v'(t) = d(v(t))/dt = d(t^2 - 9t + 18)/dt = 2t - 9

To find when v'(t) > 0, we solve the inequality:

2t - 9 > 0

t > 9/2

Since t is restricted between 0 and 8 seconds, there are no time intervals when the particle is going faster.

5. The time interval(s) when the particle is slowing down:
The particle is slowing down when the absolute value of the velocity is decreasing. The velocity function is v(t) = t^2 - 9t + 18.

To find the time intervals when the particle is slowing down, we need to look for when the velocity function is decreasing. This occurs when the derivative of the velocity function is negative:

v'(t) = d(v(t))/dt = d(t^2 - 9t + 18)/dt = 2t - 9

To find when v'(t) < 0, we solve the inequality:

2t - 9 < 0

t < 9/2

Since t is restricted between 0 and 8 seconds, the time intervals when the particle is slowing down are (0, 9/2) and (9/2, 8).

6. The total distance the particle has traveled between 0 and 8 seconds:
To find the total distance traveled, we sum the absolute values of the individual displacements. We can split the time interval into sections where the particle changes direction.

From 0 to 3 seconds, the particle moves right: s(3) - s(0) = (3^2 - 9*3 + 18) - (0^2 - 9*0 + 18) = 9 - 18 = -9 m

From 3 to 6 seconds, the particle moves left: s(6) - s(3) = (6^2 - 9*6 + 18) - (3^2 - 9*3 + 18) = 18 - 9 = 9 m

From 6 to 8 seconds, the particle moves right: s(8) - s(6) = (8^2 - 9*8 + 18) - (6^2 - 9*6 + 18) = 18 - 9 = 9 m

Now, we calculate the total distance by summing the absolute values:

Total distance = |(-9 m)| + |9 m| + |9 m| = 9 m + 9 m + 9 m = 27 m

Therefore, the total distance the particle has traveled between 0 and 8 seconds is 27 meters.

To answer each of these questions, we will first need to find the expression for the position function, s(t), by integrating the velocity function v(t). Then, we can use this position function to find the average and instantaneous velocity, speed, and determine the time intervals when the particle is moving right, going faster, slowing down, and the total distance traveled.

1. Average Velocity over the Interval 0 to 8 Seconds:
Average velocity is calculated by dividing the change in position by the change in time. In this case, since we have the position function, we will use the following formula:
Average velocity = (s(8) - s(0)) / (8 - 0)

To find s(t), integrate the velocity function v(t):
s(t) = ∫(t^2 - 9t + 18) dt
= (1/3)t^3 - (9/2)t^2 + 18t + C

Using the initial condition s(0) = 1, we can solve for C:
1 = (1/3)(0)^3 - (9/2)(0)^2 + 18(0) + C
C = 1

Thus, the position function is:
s(t) = (1/3)t^3 - (9/2)t^2 + 18t + 1

Now substitute these values into the average velocity formula to get the answer.

2. Instantaneous Velocity and Speed at Time 5 Seconds:
To find the instantaneous velocity at a specific time, we will take the derivative of the position function s(t) with respect to t and evaluate it at t = 5.
v(t) = ds(t)/dt

Take the derivative of s(t):
v(t) = d/dt [(1/3)t^3 - (9/2)t^2 + 18t + 1]
= t^2 - 9t + 18

To find the instantaneous velocity at t = 5, substitute t = 5 into v(t).

To find the speed at a specific time, we take the absolute value of the instantaneous velocity at that time.

3. Time Intervals when the Particle is Moving Right:
The particle moves right when its velocity is positive. In this case, we need to determine the intervals when v(t) > 0. By looking at the velocity function v(t) = t^2 - 9t + 18, we can observe that it is a quadratic equation that opens upward (because the coefficient of t^2 is positive).

To find the time intervals when the particle is moving right, set v(t) > 0 and solve for t.

4. Time Intervals when the Particle is Going Faster:
The particle is going faster when its velocity is increasing. This means we need to find the intervals where the velocity function is increasing, meaning its derivative is positive.

To find the time intervals when the particle is going faster, take the derivative of v(t) and solve for t when its derivative is positive.

5. Time Intervals when the Particle is Slowing Down:
The particle is slowing down when its velocity is decreasing. This means we need to find the intervals where the velocity function is decreasing, meaning its derivative is negative.

To find the time intervals when the particle is slowing down, take the derivative of v(t) and solve for t when its derivative is negative.

6. Total Distance Traveled between 0 and 8 Seconds:
Total distance is the sum of all distances traveled, regardless of direction. To find the total distance, we will integrate the absolute value of the velocity function |v(t)| over the interval t = 0 to t = 8 seconds.

Evaluate ∫|v(t)| dt from 0 to 8 to find the total distance traveled.

By following these steps, you should be able to answer each of the questions about the particle's motion.