Two particles move along the x -axis. For 0 is less than or equal to t is less than or equal to 6, the position of particle P at time t is given by p(t)=2cos((pi/4)t), while the position of particle R at time t is given by r(t)=t^3 -6t^2 +9t+3.

1. For 0 is less than or equal to t is less than or equal to 6, find all times t during which particle R is moving to the left. 2. for 0 is less than or equal to t is less than or equal to 6, find all times t during which the two particles travel in opposite directions. 3. Find the acceleration of particles P at time t=3. Is particle P speeding up, slowing down, or doing neither at time t=3? Explain your reasoning. 4. Show that during the interval (1,3), there must be at least one instant when the particle R must have a velocity of -2.

To find the answers to the given questions, we'll need to analyze the positions, velocities, and accelerations of the particles P and R over the given time interval. Let's go through each question step by step:

1. To determine when particle R is moving to the left, we need to find the intervals where the velocity of particle R is negative. The velocity of particle R is given by taking the derivative of its position function, r(t):

r'(t) = 3t^2 - 12t + 9.

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

3t^2 - 12t + 9 < 0.

Factoring the quadratic expression, we get:

(t-1)(t-3) < 0.

To solve for t, we set each factor to zero and find the intervals where the inequality holds true:

t - 1 < 0, which gives t < 1,
t - 3 < 0, which gives t < 3,

The solution is 0 ≤ t < 1.

Therefore, particle R is moving to the left for t in the interval [0, 1).

2. To find when the two particles are traveling in opposite directions, we need to consider their velocities. Particle P has a constant velocity given by the derivative of its position function, p(t):

p'(t) = -π/2 sin((π/4) t).

To find when the velocities of both particles have opposite signs, we need to determine when p'(t) and r'(t) have different signs simultaneously.

Looking at the velocity of particle R, we have:

r'(t) = 3t^2 - 12t + 9.

Comparing the signs of p'(t) and r'(t), we can determine their relationship over the interval [0, 6]:

When t < 1, p'(t) and r'(t) have opposite signs.
When 1 < t < 3, p'(t) and r'(t) have the same sign.
When t > 3, p'(t) and r'(t) have opposite signs.

Hence, the two particles travel in opposite directions for t in the intervals (0, 1) U (3, 6].

3. To find the acceleration of particle P at time t = 3, we need to differentiate the velocity function of particle P:

p''(t) = (-π/2)((π/4) cos((π/4) t)) = -π^2/8 cos((π/4) t).

Evaluating p''(3), we get:

p''(3) = -π^2/8 cos((3π/4)) = -π^2/8 (-√2/2) = π^2/16.

The acceleration of particle P at t = 3 is π^2/16.

To determine if particle P is speeding up, slowing down, or doing neither at t = 3, we analyze the sign of p''(t). Since p''(t) is positive (as π^2/16 is positive), particle P is accelerating or speeding up at t = 3.

4. To prove that particle R must have a velocity of -2 at least once during the interval (1, 3), we need to analyze the velocity function of particle R:

r'(t) = 3t^2 - 12t + 9.

The task is to show that r'(t) = -2 has a root between t = 1 and t = 3.

By substituting -2 for r'(t), we get:

3t^2 - 12t + 9 = -2.

Rearranging the equation, we have:

3t^2 - 12t + 11 = 0.

Using the quadratic formula, we find:

t = (12 ± √(12^2 - 4(3)(11))) / (2(3)).

Solving for t, we get two values: approximately t ≈ 1.72 and t ≈ 2.28.

Since both t = 1.72 and t = 2.28 are within the interval (1, 3), we conclude that particle R must have a velocity of -2 at least once during the interval (1, 3).

Note: To find the exact values of t, you can use numerical methods or a graphing calculator.