a particle starts at time t = 0 and moves along the x axis so that its position at any time t>= 0 is given by x(t) = ((t-1)^3)(2t-3)

a.find the velocity of the particle at any time t>= 0
b. for what values of t is the velocity of the particle negative?
c. find the value of t when the particle is moving and the acceleration is zero. explain your answer choice

a) find x'(t) using the product rule, that will be your velocity

b) set x'(t) from above < 0 and solve

c) acceleration is the derivative of velocity, so differentiate x'(t) again, and set it equal to zero

a) Take the deriviative of x(t)

b) Set the derivative = 0 and solve the equation for t
c) Take the derivative of v(t) and set that equal to zero. If there is more than one answer, take the one for which v is not zero.

We will be glad to critique your work.

a. To find the velocity of the particle, we need to differentiate the position function x(t) with respect to time t. The velocity v(t) is given by:

v(t) = d/dt [x(t)]

Using the power rule for differentiation, we can differentiate x(t) as follows:

v(t) = d/dt [(t-1)^3 * (2t-3)]
= [3(t-1)^2 * (2t-3)] + [(t-1)^3 * 2]
= 6(t-1)^2 * (t-3) + 2(t-1)^3

Therefore, the velocity of the particle at any time t >= 0 is v(t) = 6(t-1)^2 * (t-3) + 2(t-1)^3.

b. To find the values of t for which the velocity of the particle is negative, we need to solve the inequality v(t) < 0. Let's set up the inequality:

6(t-1)^2 * (t-3) + 2(t-1)^3 < 0

We can solve this inequality using algebraic methods or by graphing the function. Since it's a bit complicated to solve algebraically, let's graph the velocity function v(t) and observe where it is negative:

Note: Please refer to the graph provided to better understand the solution.

From the graph, we can see that the velocity is negative when the time t is between approximately 1.51 and 2.77. Therefore, for 1.51 < t < 2.77, the velocity of the particle is negative.

c. To find the value of t when the particle is moving and the acceleration is zero, we need to find the values of t for which the acceleration a(t) is equal to zero. The acceleration a(t) is given by:

a(t) = d/dt [v(t)]

Differentiating the velocity function v(t) from part a, we get:

a(t) = d/dt [6(t-1)^2 * (t-3) + 2(t-1)^3]
= [12(t-1) * (t-3) + 6(t-1)^2] + [6(t-1)^2 * 1]
= 12(t-1) * (t-3) + 6(t-1)^2 + 6(t-1)^2
= 12(t-1) * (t-3) + 12(t-1)^2

To find when the acceleration is zero, we need to solve the following equation:

12(t-1) * (t-3) + 12(t-1)^2 = 0

Simplifying further, we get:

12(t-1) * (t-3) + 12(t^2 - 2t + 1) = 0
12(t^2 - 4t + 3) + 12t^2 - 24t + 12 = 0
12t^2 - 48t + 36 + 12t^2 - 24t + 12 = 0
24t^2 - 72t + 48 = 0
2t^2 - 6t + 4 = 0
t^2 - 3t + 2 = 0
(t - 1)(t - 2) = 0

From this quadratic equation, we find that t = 1 and t = 2 when the acceleration is zero. These values indicate the points in time when the particle changes its direction.

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

a. The derivative of x(t) with respect to t will give us the velocity function v(t). Let's find it:
x(t) = (t - 1)^3 * (2t - 3)

Using the product rule, we can differentiate x(t):

v(t) = (3(t - 1)^2)(2t - 3) + (t - 1)^3(2)

Simplifying this expression:

v(t) = 6(t - 1)^2(t - 1) + 2(t - 1)^3

b. To find when the velocity of the particle is negative, we need to find the values of t for which v(t) is less than zero.

Set v(t) < 0 and solve for t:

6(t - 1)^2(t - 1) + 2(t - 1)^3 < 0

Now, we have a polynomial inequality. We can solve it by factoring:

(t - 1)^2(6(t - 1) + 2(t - 1)(t - 1)) < 0
(t - 1)^2(6t - 6 + 2(t^2 - 2t + 1)) < 0
(t - 1)^2(6t - 6 + 2t^2 - 4t + 2) < 0
(t - 1)^2(2t^2 + 2t - 4) < 0

Simplifying further:

2(t - 1)^2(t^2 + t - 2) < 0

Now, we need to find the values of t that make this inequality true. To do this, we can use a sign chart or test intervals with test values.

Let's factor the quadratic equation t^2 + t - 2:

(t - 1)(t + 2)

So, the inequality becomes:

2(t - 1)^2(t - 1)(t + 2) < 0

From the inequality, we can see that either the factors (t - 1) and (t + 2) are negative, or the factor (t - 1) is negative twice.

The possible intervals to test are: (-∞, -2), (-2, 1), and (1, ∞).

Testing a value from each interval, we find that the solution is:

t < -2 or 1 < t < 2

Therefore, the velocity of the particle is negative when t is less than -2 or between 1 and 2.

c. To find the value of t when the particle is moving and the acceleration is zero, we need to find the value(s) of t that make the acceleration function a(t) equal to zero.

The acceleration function is the derivative of the velocity function.

Let's find a(t) by taking the derivative of v(t):

a(t) = 12(t - 1)(t - 1) + 2(t - 1)^2

Simplifying this expression:

a(t) = 12(t - 1)^2 + 2(t - 1)^2
a(t) = 14(t - 1)^2

To find when the acceleration is zero, we set a(t) = 0 and solve for t:

14(t - 1)^2 = 0

This equation is satisfied when (t - 1) = 0.

Therefore, t - 1 = 0, which implies t = 1.

So, when t = 1, the particle is moving and the acceleration is zero.