To solve this problem, we need to tackle both parts:

Given: $v(t) = t^3 - 12t^2 + 41t - 30 \quad \text{for} \quad 1 \leq t \leq 7$

(a) Find the displacement.

Displacement is the net change in position, which is the integral of the velocity function over the interval $[1, 7]$:

$\text{Displacement} = \int_{1}^{7} v(t) \, dt$

(b) Find the total distance.

The total distance is the integral of the absolute value of the velocity over the interval $[1, 7]$. This means we need to identify where $v(t) = 0$ within $[1, 7]$ to determine any intervals where the velocity is negative, as the particle might reverse direction.

To proceed with these calculations, let's:

1. Integrate $v(t)$ from $1$ to $7$ for displacement.
2. Find the points where $v(t) = 0$ within the interval and set up the integral for total distance by breaking up the interval based on these points, ensuring we take the absolute value of $v(t)$ over the intervals.

Let me compute these integrals for a detailed solution.The solutions for the problem are as follows:

(a) Displacement

The displacement of the particle over the interval $[1, 7]$ is: $36 \text{ feet}$

(b) Total Distance

The total distance traveled by the particle over the interval $[1, 7]$ is: $37.5 \text{ feet}$

Would you like a breakdown of the integration process or have any questions?

Here are some related questions to deepen your understanding:

1. How do you interpret the difference between displacement and total distance?
2. What would happen to the displacement and distance values if the interval was different?
3. How can you find points where velocity changes direction in other problems?
4. What if the velocity function had an additional trigonometric term? How would the approach change?
5. Can this method be applied to acceleration functions to find velocity over time?

Tip: Remember, displacement considers direction (net change in position), while total distance is always a positive sum of all movement, regardless of direction changes.