Let's go through the problem.

We are given the function for the position of a particle $s(t)$ in terms of time $t$:

$f(t) = \sin\left( \frac{\pi t}{2} \right)$

(a) Find the velocity $v(t)$ at time $t$:

The velocity is the derivative of the position function. Thus, we need to differentiate $f(t) = \sin\left( \frac{\pi t}{2} \right)$ with respect to $t$.

$v(t) = \frac{d}{dt} \left[ \sin\left( \frac{\pi t}{2} \right) \right]$

Using the chain rule, the derivative of $\sin(u)$ is $\cos(u)$, and the derivative of $u = \frac{\pi t}{2}$ is $\frac{\pi}{2}$:

$v(t) = \frac{\pi}{2} \cos\left( \frac{\pi t}{2} \right)$

So the velocity function is:

$v(t) = \frac{\pi}{2} \cos\left( \frac{\pi t}{2} \right)$

(b) Find the velocity at $t = 1$:

To find the velocity after 1 second, substitute $t = 1$ into the velocity function:

$v(1) = \frac{\pi}{2} \cos\left( \frac{\pi (1)}{2} \right)$

$v(1) = \frac{\pi}{2} \cos\left( \frac{\pi}{2} \right)$

Since $\cos\left( \frac{\pi}{2} \right) = 0$:

$v(1) = \frac{\pi}{2} \times 0 = 0$

Thus, the velocity at $t = 1$ is $0 \, \text{ft/s}$.

(c) When is the particle at rest?

The particle is at rest when the velocity is zero. Therefore, set the velocity function equal to zero and solve for $t$:

$\frac{\pi}{2} \cos\left( \frac{\pi t}{2} \right) = 0$

Since $\frac{\pi}{2} \neq 0$, we must have:

$\cos\left( \frac{\pi t}{2} \right) = 0$

The cosine function equals zero at $\frac{\pi t}{2} = \frac{\pi}{2} + n\pi$, where $n$ is an integer. Solving for $t$:

$\frac{\pi t}{2} = \frac{\pi}{2} + n\pi$

$t = 1 + 2n$

Thus, the particle is at rest at $t = 1 + 2n$, where $n$ is any integer.

Let me know if you need more details or clarifications!

Here are 5 questions to expand on this:

1. What is the particle’s position at $t = 0$?
2. What is the acceleration function of the particle?
3. At what other times (besides $t = 1$) is the velocity zero?
4. How does the velocity behave as $t \to \infty$?
5. Can you interpret the motion of the particle based on the position and velocity functions?

Tip: The derivative of a trigonometric function often requires the chain rule. Be sure to differentiate the inner function as well!