To solve this problem, you need to find the times at which the particle is at rest and changes direction from left to right.
The velocity of the particle is given by the derivative of the position function: v(t) = 3cos(2t) + 2sin(t).
The particle is at rest when its velocity is zero, so we need to solve the equation v(t) = 0. This gives us:
3cos(2t) + 2sin(t) = 0
To solve this equation, you can try setting each of the trigonometric functions equal to zero and solving for t. This gives:
3cos(2t) = 0 => t = pi/4, 5pi/4, 9pi/4, 13pi/4, ...
2sin(t) = 0 => t = 0, pi, 2pi, 3pi, ...
The second solution gives all the times at which the particle is at rest. The first time the particle is at rest is at t = 0, and the second time the particle is at rest is at t = 2pi, which is approximately 2.222. Therefore, the answer to the first question is (D) 2.222.
To find the times at which the particle changes direction from left to right, we need to find the times at which the velocity changes sign. The velocity changes sign when it goes from positive to negative or from negative to positive.
The velocity is positive when cos(2t) is positive and sin(t) is positive. The velocity is negative when cos(2t) is negative and sin(t) is negative.
You can use a sign chart to help you determine the signs of the trigonometric functions at different values of t:
t 0 pi/4 pi/2 3pi/4 pi 5pi/4 3pi/2 7pi/4 2pi
cos(2t) 1 -0.5 0 -0.5 1 -0.5 0 -0.5 1
sin(t) 0 1 1 0 0 -1 -1 0 0