Question

A particle moves along the x-axis so that its velocity at any time t ≥ 0 is given by

v(t) = (2(pi) − 5)t − sin(t(pi))
A. Find the acceleration at any time t.
B. Find the minimum acceleration of the particle over the interval [0, 3].
C. Find the maximum velocity of the particle over the interval [0, 2].

Answer 1

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.