5. A particle moves along the x-axis in such a way that its position at time t is given by x=3t^4-16t^3+24t^2 for -5 ≤ t ≤ 5.

a. Determine the velocity and acceleration of the particle at time t.
b. At what values of t is the particle at rest?
c. At what values of t does the particle change direction?
d. What is the velocity when the acceleration is first zero?

a) velocity = x' = 12t^3 - 48t^2 + 48t

acc. = x'' = 36t^2 - 96t + 48

b) to be at rest, x' = 0
12t^3 - 48t^2 + 48t = 0
12t(t^2 - 4t + 4) = 0
12t(t-2)^2 = 0
t = 0 or t = 2

c) mmmh, what must have happened to the velocity when direction is changed?

d) for acc. = 0, 36t^2 - 96t + 48 = 0
3t^2 - 8t + 4 = 0
(t-2)(3t - 2) = 0
t = 2 or t = 2/3

so when t = 2/3, sub that into x'
I will let you do the arithmetic.

a. To find the velocity, we need to take the derivative of the position function with respect to time.

Position function: x = 3t^4 - 16t^3 + 24t^2

Velocity function: v = d/dt (x) = d/dt (3t^4 - 16t^3 + 24t^2)

Taking the derivative, we get:

v = 12t^3 - 48t^2 + 48t

To find the acceleration, we need to take the derivative of the velocity function with respect to time.

Acceleration function: a = d/dt (v) = d/dt (12t^3 - 48t^2 + 48t)

Taking the derivative, we get:

a = 36t^2 - 96t + 48

b. The particle is at rest when its velocity is zero. Set the velocity function (v) equal to zero and solve for t:

0 = 12t^3 - 48t^2 + 48t

Factor out common terms:

0 = 12t(t^2 - 4t + 4)

Simplify further:

0 = 12t(t - 2)^2

Setting each factor equal to zero, we get:

12t = 0 or t - 2 = 0
t = 0 t = 2

So, the particle is at rest at t = 0 and t = 2.

c. The particle changes direction when the velocity changes sign. In other words, when the velocity is positive and then becomes negative or when the velocity is negative and then becomes positive.

To determine when the velocity changes sign, we look at the intervals where the velocity is positive and negative. We can do this by finding the critical points of the velocity function (v) and testing values within each interval.

For v = 12t^3 - 48t^2 + 48t:

Set v = 0 and solve for t to find the critical points:

12t^3 - 48t^2 + 48t = 0

Factor out common terms:

12t(t^2 - 4t + 4) = 0

Simplify further:

12t(t - 2)^2 = 0

Setting each factor equal to zero, we get:

12t = 0 or t - 2 = 0
t = 0 t = 2

These critical points divide the time interval into three intervals: -5 ≤ t < 0, 0 < t < 2, and 2 < t ≤ 5.

Testing values within each interval:

For -5 ≤ t < 0: Choose t = -1

v = 12(-1)^3 - 48(-1)^2 + 48(-1) = -12 + 48 - 48 = -12

Since the velocity is negative, the particle is moving to the left in this interval.

For 0 < t < 2: Choose t = 1

v = 12(1)^3 - 48(1)^2 + 48(1) = 12 - 48 + 48 = 12

Since the velocity is positive, the particle is moving to the right in this interval.

For 2 < t ≤ 5: Choose t = 4

v = 12(4)^3 - 48(4)^2 + 48(4) = 192 - 768 + 192 = -384

Since the velocity is negative, the particle is moving to the left in this interval.

From the analysis, the particle changes direction at t = 0 and t = 2.

d. The velocity is zero when the particle is momentarily at rest. We can find the velocity by plugging in the critical points into the velocity function:

For t = 0:

v = 12(0)^3 - 48(0)^2 + 48(0) = 0

For t = 2:

v = 12(2)^3 - 48(2)^2 + 48(2) = 96 - 192 + 96 = 0

So, the velocity when the acceleration is first zero is zero.

To determine the velocity and acceleration of the particle at time t, we need to take the derivative of the position function x(t).

a. Velocity: To find the velocity, we take the derivative of x(t) with respect to t.

The derivative of x(t) with respect to t is:
v(t) = d/dt(3t^4-16t^3+24t^2)

To differentiate each term, we use the power rule for derivatives.

v(t) = 12t^3 - 48t^2 + 48t

b. Acceleration: To find the acceleration, we take the derivative of velocity function v(t).

The derivative of v(t) with respect to t is:
a(t) = d/dt(12t^3 - 48t^2 + 48t)

Again, we use the power rule for derivatives to differentiate each term.

a(t) = 36t^2 - 96t + 48

Now, let's move on to the next parts of the question.

b. Particle at rest: The particle is at rest when the velocity is zero.

To find the values of t when the particle is at rest, we set v(t) = 0:

0 = 12t^3 - 48t^2 + 48t

To solve this equation, we can factor out common terms:

0 = 12t(t^2 - 4t + 4)

Since we want the particle to be at rest, t = 0 is one solution.

To find the other solutions, we solve the quadratic equation t^2 - 4t + 4 = 0:

(t - 2)(t - 2) = 0

So, t = 2 is the other value when the particle is at rest.

c. Change in direction: The particle changes direction when the velocity changes sign.

To find the values of t when the particle changes direction, we look for the zeros of v(t):

12t^3 - 48t^2 + 48t = 0

We can factor out common terms:

12t(t^2 - 4t + 4) = 0

And we can factor the quadratic equation:

(t - 2)(t - 2) = 0

Again, t = 2 is the value when the particle changes direction.

d. Velocity when acceleration is first zero: To find the velocity when the acceleration is first zero, we set a(t) = 0:

0 = 36t^2 - 96t + 48

We can divide the equation by 12 to simplify it:

0 = 3t^2 - 8t + 4

To solve this quadratic equation, we can factor it:

0 = (3t - 2)(t - 2)

So, the velocity is first zero when t = 2/3 and t = 2.

By following these procedures, we have determined the velocity, acceleration, points of rest, changes in direction, and the velocity when the acceleration is first zero.