The velocity of a particle moving along the x-axis at any time t >= 0 is given by v(t) = cos(pi t) - t(7 - 2pi).

a. Find the acceleration at any time t.

b Find the particle's maximum acceleration over the interval [0, 2]. Use derivatives to support and justify your answer.

c. Find the particle's minimum velocity over the interval [0, 3]. Justify your answer.

Please show all work, all help is greatly appreciated.

v(t) = cos(pi t) - t(7 - 2pi)

Please check for typo in t(7 - 2pi)
As written it is not part of your problem.
a = dv/dt = - pi sin (pi t)

when is a max?
when da/dt = 0
da/dt = -pi^2 cos pi t = 0 at max
cos is 0 when pi t = pi/2 and when pi t = 3 pi/2 and v is positive
but at t = 1/2 a is negative
so at t = 3/2
a max = - pi sin (pi t)= - pi sin (3pi/2) = - pi(-1) = +pi at max

do c the same way. velocity is min when a = zero and v is negative

Unless that was not a typo and your velocity keeps diving forever due to that weird

t(7 - 2pi)

That's not a typo, that's just how the question is written.

a. To find the acceleration at any time t, we need to take the derivative of the velocity function v(t) with respect to time t. The derivative of cos(pi t) is -pi sin(pi t), and the derivative of -t(7 - 2pi) is -(7 - 2pi).

So, the acceleration function a(t) is given by:

a(t) = -pi sin(pi t) - (7 - 2pi)

b. To find the particle's maximum acceleration over the interval [0, 2], we need to find the critical points of the acceleration function a(t) within this interval.

The critical points occur where the derivative of a(t) equals zero or does not exist. Let's find the derivative of a(t):

a'(t) = -pi^2 cos(pi t)

Setting this equal to zero, we have:

-pi^2 cos(pi t) = 0

Since cos(pi t) can never be zero, we conclude that there are no critical points within the interval [0, 2].

To determine the maximum acceleration, we need to evaluate a(t) at the endpoints of the interval [0, 2].

a(0) = -pi sin(0) - (7 - 2pi) = - (7 - 2pi) = 2pi - 7

a(2) = -pi sin(2pi) - (7 - 2pi) = 0 - (7 - 2pi) = 2pi - 7

Therefore, the maximum acceleration over the interval [0, 2] is 2pi - 7.

c. To find the particle's minimum velocity over the interval [0, 3], we need to find the critical points of the velocity function v(t) within this interval.

The critical points occur where the derivative of v(t) equals zero or does not exist. Let's find the derivative of v(t):

v'(t) = -pi sin(pi t) - (7 - 2pi)

Setting this equal to zero, we have:

-pi sin(pi t) - (7 - 2pi) = 0

Rearranging the equation, we have:

-pi sin(pi t) = 7 - 2pi

Dividing by -pi, we get:

sin(pi t) = (7 - 2pi) / pi

To find the minimum velocity, we need to evaluate v(t) at the critical points and the endpoints of the interval [0, 3]. However, solving the equation above for t to find the critical points is quite challenging algebraically.

Since analyzing the derivative and finding the critical points is complex, another approach to finding the minimum velocity is to evaluate the velocity function v(t) at the endpoints of the interval [0, 3].

v(0) = cos(pi * 0) - 0(7 - 2pi) = 1

v(3) = cos(pi * 3) - 3(7 - 2pi) = -1 - 21 + 6pi

To compare these values, we can rewrite v(3) as -22 + 6pi.

Since -1 - 21 + 6pi is less than 1 for any value of pi, we can conclude that the minimum velocity occurs at t = 3 within the interval [0, 3].

Therefore, the particle's minimum velocity over the interval [0, 3] is -22 + 6pi.