To find the distance traveled by the particle over the 24-second interval, we can use the velocity function $v(t)$ and integrate it with respect to time to get the displacement.

The velocity function is given as:

$v(t) = \frac{1}{288} t (576 - t^2)$

To find the distance traveled from $t = 0$ to $t = 24$, we integrate $v(t)$ over this interval:

$s = \int_{0}^{24} v(t) \, dt = \int_{0}^{24} \frac{1}{288} t (576 - t^2) \, dt$

Let's compute the integral step by step.

1. Expand the expression inside the integral:

$v(t) = \frac{1}{288} t (576 - t^2) = \frac{1}{288} (576t - t^3)$

Now the integral becomes:

$s = \frac{1}{288} \int_{0}^{24} (576t - t^3) \, dt$

2. Split the integral:

$s = \frac{1}{288} \left[ \int_{0}^{24} 576t \, dt - \int_{0}^{24} t^3 \, dt \right]$

3. Integrate each term:

• The integral of $576t$ is:

$\int 576t \, dt = 288t^2$

Evaluating this from 0 to 24 gives:

$288(24^2) - 288(0^2) = 288(576) = 165888$

• The integral of $t^3$ is:

$\int t^3 \, dt = \frac{t^4}{4}$

Evaluating this from 0 to 24 gives:

$\frac{24^4}{4} - \frac{0^4}{4} = \frac{331776}{4} = 82944$

4. Combine the results:

Now, subtract the integrals and multiply by $\frac{1}{288}$:

$s = \frac{1}{288} \left( 165888 - 82944 \right) = \frac{1}{288} (82944) = 288$

Thus, the distance traveled by the particle in 24 seconds is 288 units.

---

Would you like any further clarifications or additional details? Here are five related questions:

1. What is the difference between displacement and distance traveled?
2. How does integrating velocity give the displacement?
3. How would the result change if the velocity function had a different form?
4. Can this technique be applied to a velocity function with trigonometric terms?
5. How do we interpret negative velocity in terms of displacement?

Tip: When velocity changes direction, be mindful of whether you're asked for total distance (which is always positive) or displacement (which can be negative).