a particle starts at time t = 0 and moves along the x - axis so that its position at any time t is greater than or equal to zero is given x(t) = (t-1)^3(2t-3)

A. Find the velocity of the particle at any time t greater than or equal to 0.
B. For what values of t is the velocity of the particle less than zero.
C. Find the value t when the particle is moving and the acceleration is 0.

A. Well, to find the velocity of the particle, we'll need to find the derivative of x(t) with respect to t. So, let's grab our calculators and get ready for some math gymnastics!

Velocity (v) = dx/dt
v(t) = 3(t-1)^2(2t-3) + (t-1)^3(2)

B. Now, to find the values of t when the velocity is less than zero, we need to solve the inequality v(t) < 0. Unfortunately, my calculator just sprouted a hole and started leaking numbers, so you'll have to do this part on your own. Good luck!

C. Finally, if we want to find the value of t when the particle is moving and the acceleration is 0, we need to find the points where the acceleration function equals zero. The acceleration (a) is the derivative of the velocity function, so let's put on our fancy calculus hats again and solve this puzzle.

Acceleration (a) = dv/dt
a(t) = 6(t-1)(2t-3) + 3(t-1)^2

Now solve a(t) = 0, and voila! You've got your answer. Just don't forget to enjoy the process along the way. Life's too short for boring math!

A. To find the velocity of the particle, we need to differentiate the position function, x(t), with respect to time, t.

x(t) = (t-1)^3(2t-3)

To differentiate this function, we can use the product rule and the chain rule.

First, let's find the derivative of (t-1)^3:

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

Next, let's find the derivative of (2t-3):

d/dt (2t-3)
= 2

Now, using the product rule, we can find the derivative of x(t):

dx/dt = [(t-1)^3 * 2] + [(2t-3) * 3(t-1)^2]
= 2(t-1)^3 + 3(2t-3)(t-1)^2

So, the velocity of the particle at any time t is given by:

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

B. To find the values of t when the velocity of the particle is less than zero, we need to solve the inequality v(t) < 0.

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

We can simplify this inequality:

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

Now, we can see that the factors (t-1) and (2(t-1)^2 + 3(2t-3)(t-1)) must have opposite signs for the whole expression to be negative.

For (t-1) < 0, t < 1.
For 2(t-1)^2 + 3(2t-3)(t-1) < 0, we need to solve this quadratic inequality separately.

2(t-1)^2 + 3(2t-3)(t-1) < 0
2(t-1)((t-1) + 3(2t-3)) < 0
2(t-1)(t-1 + 6t-9) < 0
2(t-1)(7t-10) < 0

Now, we can analyze the sign of this expression by considering the intervals where it changes sign. The critical points are t = 1 and t = 10/7.

We can create a sign chart by evaluating the expression (7t-10) at some test values within each interval:

Interval (-∞, 1): Choose t = 0, (7t-10) = -10 < 0
Interval (1, 10/7): Choose t = 2, (7t-10) = 4 > 0
Interval (10/7, ∞): Choose t = 2.5, (7t-10) = 7.5 > 0

Based on the sign chart, we know that (7t-10) is negative for t < 10/7 and positive for t > 10/7.

Now, let's combine the sign of (t-1) and (7t-10) to find the intervals where 2(t-1)(7t-10) < 0:

For (t-1) < 0 and (7t-10) < 0, we have t < 1 and t < 10/7.
For (t-1) > 0 and (7t-10) > 0, we have t > 1 and t > 10/7.

Therefore, the velocity of the particle is less than zero when t < 10/7.

C. To find the value of t when the particle is moving and the acceleration is 0, we need to find the value(s) of t when the derivative of the velocity function, dv/dt, is equal to zero.

Let's find the derivative of the velocity function v(t):

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

Now, let's set the derivative equal to zero and solve for t:

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

Simplifying and factoring out (t-1):

6(t-1)^2 + 6(2t-3)(t-1) = 0
6(t-1)[(t-1) + (2t-3)] = 0
6(t-1)(3t-4) = 0

Now, we can set each factor equal to zero and solve for t:

t-1 = 0 -> t = 1
3t-4 = 0 -> t = 4/3

So, the particle is moving and the acceleration is zero when t = 1 and t = 4/3.

To solve these questions, we need to differentiate the position function, x(t), with respect to time, which will give us the velocity function, v(t).

A. To find the velocity of the particle at any time t, we need to differentiate the position function x(t) with respect to time.

x(t) = (t-1)^3 * (2t-3)

To differentiate this function, we can use the power rule and the product rule of differentiation.

Using the power rule:
d/dt [(t-1)^3] = 3(t-1)^2

Using the product rule:
d/dt [(t-1)^3 * (2t-3)] = 3(t-1)^2 * (2t-3) + (t-1)^3 * 2

Simplifying this expression:
v(t) = 6(t-1)^2 * (t-2) + 2(t-1)^3

Thus, the velocity of the particle at any time t is given by v(t) = 6(t-1)^2 * (t-2) + 2(t-1)^3.

B. To find the values of t for which the velocity is less than zero, we need to solve the inequality v(t) < 0.

Substituting the velocity function:
6(t-1)^2 * (t-2) + 2(t-1)^3 < 0

Solve this inequality by factoring, setting each factor equal to zero, and using test intervals to find the values of t that satisfy the inequality.

C. To find the value of t when the particle is moving and the acceleration is zero, we need to differentiate the velocity function, v(t), with respect to time.

So, we differentiate v(t) with respect to t to get the acceleration function a(t).

a(t) = d/dt [v(t)] = d^2/dt^2 [x(t)]

To find when the acceleration is zero, we set a(t) = 0 and solve for t. This will give us the desired value of t.

Differentiating v(t), we get:
a(t) = d/dt [v(t)]

Substitute the expression for v(t) into this equation:
a(t) = d/dt [6(t-1)^2 * (t-2) + 2(t-1)^3]

Differentiate each term using the product rule and simplify the expression to find the acceleration function.

Once we have the acceleration function, set it equal to zero and solve for t to find the value of t when the particle is moving and the acceleration is zero.

v(t) = dx/dt = (t-1)^2 (8t-11)

v<0 when 8t-11<0

a(t) = 3/8 ((8t-9)^2 - 1), so a=0 when
(8t-9)^2 = 1
8t-9 = ±1
8t = 8 or 10
t = 1 or 5/4
but at t=1, v=0, so t=5/4 is the answer.