The position of a particle moving on a horizontal line is given by s(t)=2t^3-15t^2+24t-5, where s is measured in feet and t in seconds.

a: What is the initial position of the particle?
b: What is the average velocity of the particle on the interval [0,2]?Indicate units of measure
c: Find a formula for the instantaneous velocity of the particle.
d: When is the particle at rest?
e: When is the particle moving to the right? When is it moving to the left?
f: At t=2 seconds, is the particle moving away from the origin or towards the origin?
h: What is the average acceleration of the particle on the interval [0,2]?
i: Find a formula for the instantaneous acceleration of the particle.
j: At t=2 seconds, is the particle slowing down or speeding up?

a) sub in t = 0

b) avg vel = (s(2) - s(0) )/(2-0) = ...
c) s ' (t) = 6t^2 - 30t + 24

d) particle is at rest, when velocity = 0 , that is ...
6t^2 - 30t + 24 = 0
t^2 - 5t +4 = 0
(t-1)(t-4) = 0
when t = 1 or when t = 4

f) sub t=2 into the derivative of c)
if it is positive, ---> to the right
if it is negative, ---> to the left

e) same steps as b) except use the velocity expression

you do some of them.

How do you do E on this problem??

a: To find the initial position of the particle, we need to evaluate the position function at time t=0.

Substituting t=0 into the position function:
s(0) = 2(0)^3 - 15(0)^2 + 24(0) - 5 = -5

Therefore, the initial position of the particle is -5 feet.

b: The average velocity of the particle on the interval [0,2] is given by the change in position divided by the change in time.
The change in position is s(2) - s(0) and the change in time is 2 - 0.

Substituting t=2 into the position function:
s(2) = 2(2)^3 - 15(2)^2 + 24(2) - 5 = 17

Therefore, the average velocity on the interval [0,2] is:
(17 - (-5)) / (2 - 0) = 22/2 = 11 feet per second.

c: The instantaneous velocity of the particle is given by the derivative of the position function.
Taking the derivative of s(t):
v(t) = 6t^2 - 30t + 24

Therefore, the formula for the instantaneous velocity of the particle is v(t) = 6t^2 - 30t + 24 feet per second.

d: The particle is at rest when its velocity is equal to zero.
Setting the instantaneous velocity equal to zero and solving for t:
6t^2 - 30t + 24 = 0

Factoring the quadratic equation:
2(3t^2 - 15t + 12) = 0
3t^2 - 15t + 12 = 0

Solving the quadratic equation:
(t - 1)(3t - 12) = 0
t = 1 or t = 4/3

Therefore, the particle is at rest at t = 1 second and t = 4/3 seconds.

e: The particle is moving to the right when its velocity is positive and moving to the left when its velocity is negative.
By looking at the formula for the instantaneous velocity, we can see that the particle is moving to the right when t > 2 and moving to the left when t < 2.

f: To determine if the particle is moving away from or towards the origin at t=2 seconds, we need to look at the sign of its velocity.
Substituting t=2 into the instantaneous velocity formula:
v(2) = 6(2)^2 - 30(2) + 24 = 12 - 60 + 24 = -24

Since the velocity is negative, the particle is moving towards the origin at t=2 seconds.

h: The average acceleration of the particle on the interval [0,2] is given by the change in velocity divided by the change in time.
The change in velocity is v(2) - v(0) and the change in time is 2 - 0.

Substituting t=2 into the instantaneous velocity formula:
v(2) = 6(2)^2 - 30(2) + 24 = 12 - 60 + 24 = -24

Therefore, the average acceleration on the interval [0,2] is:
(-24 - 24) / (2 - 0) = -48/2 = -24 feet per second squared.

i: The instantaneous acceleration of the particle is given by the derivative of the instantaneous velocity function.
Taking the derivative of v(t):
a(t) = 12t - 30

Therefore, the formula for the instantaneous acceleration of the particle is a(t) = 12t - 30 feet per second squared.

j: To determine if the particle is slowing down or speeding up at t=2 seconds, we need to look at the sign of its acceleration.
Substituting t=2 into the instantaneous acceleration formula:
a(2) = 12(2) - 30 = 24 - 30 = -6

Since the acceleration is negative, the particle is slowing down at t=2 seconds.

a: To find the initial position of the particle, we need to find s(0) - the value of the position function when t = 0.

Substituting t = 0 into the given position function, we have:
s(0) = 2(0)^3 - 15(0)^2 + 24(0) - 5
= 0 - 0 + 0 - 5
= -5
Therefore, the initial position of the particle is -5 feet.

b: To find the average velocity of the particle on the interval [0,2], we need to find the change in position divided by the change in time.
The change in position is given by s(2) - s(0), and the change in time is 2 - 0.
Substituting t = 2 and t = 0 into the given position function, we have:
s(2) = 2(2)^3 - 15(2)^2 + 24(2) - 5
= 16 - 60 + 48 - 5
= -1
s(0) = -5
Therefore, the average velocity is (s(2) - s(0)) / (2 - 0) = (-1 - (-5)) / 2 = 4 / 2 = 2 feet per second.

c: The instantaneous velocity of the particle at any given time t is given by the derivative of the position function s(t). To find the instantaneous velocity, we need to differentiate the position function with respect to t.

d: To find when the particle is at rest, we need to find the values of t for which the instantaneous velocity is 0. In other words, we need to find the roots of the equation for the instantaneous velocity.

e: To determine when the particle is moving to the right or moving to the left, we can examine the sign of the instantaneous velocity. If the instantaneous velocity is positive, the particle is moving to the right, and if it is negative, the particle is moving to the left.

f: To determine if the particle is moving away from the origin or towards the origin at t = 2 seconds, we can examine the sign of the instantaneous velocity at t = 2. If the instantaneous velocity is positive, the particle is moving away from the origin, and if it is negative, the particle is moving towards the origin.

h: To find the average acceleration of the particle on the interval [0,2], we need to find the change in velocity divided by the change in time. The change in velocity is given by v(2) - v(0), and the change in time is 2 - 0. We need to find the formula for the instantaneous velocity, v(t), in order to calculate the average acceleration.

i: The instantaneous acceleration of the particle at any given time t is given by the derivative of the velocity function v(t). To find the instantaneous acceleration, we need to differentiate the velocity function with respect to t.

j: To determine if the particle is slowing down or speeding up at t = 2 seconds, we can examine the sign of the instantaneous acceleration at t = 2. If the instantaneous acceleration is positive, the particle is speeding up, and if it is negative, the particle is slowing down.