Answer:
A. To find the acceleration, we need to take the derivative of the velocity function with respect to time:
a(t) = v'(t) = 2(pi) - cos(t(pi))
B. To find the minimum acceleration, we need to find the critical points of the acceleration function in the interval [0, 3].
a'(t) = sin(t(pi))
The critical points occur when sin(t(pi)) = 0, which means t = 0, 1, 2, 3. We need to evaluate the acceleration function at these points and at the endpoints of the interval:
a(0) = 2(pi) - cos(0) = 2(pi)
a(1) = 2(pi) - cos(pi) = pi + 2
a(2) = 2(pi) - cos(2pi) = 2(pi)
a(3) = 2(pi) - cos(3pi) = pi - 2
The minimum acceleration occurs at t = 3, with a minimum value of pi - 2.
C. To find the maximum velocity, we need to find the critical points of the velocity function in the interval [0, 2].
v'(t) = 2(pi) - cos(t(pi)) = 0
The critical points occur when cos(t(pi)) = 2(pi). We can solve for t as follows:
cos(t(pi)) = 2(pi)
t(pi) = arccos(2(pi))
t = arccos(2(pi))/pi ≈ 1.58
We need to evaluate the velocity function at these points and at the endpoints of the interval:
v(0) = -sin(0) = 0
v(1.58) ≈ 1.69
v(2) = (2(pi) - 5)(2) - sin(2(pi)) = 4(pi) - 10
The maximum velocity occurs at t = 1.58, with a maximum value of approximately 1.69.