To find the total distance traveled by the particle during the first $t$ seconds, we need to integrate the velocity function $v(t) = Ze^{-2t}$ over the interval from $t = 0$ to $t = t$.

The distance $D(t)$ is given by:

$D(t) = \int_0^t v(t) \, dt = \int_0^t Ze^{-2t} \, dt$

Now, we solve the integral:

$D(t) = Z \int_0^t e^{-2t} \, dt$

To integrate $e^{-2t}$, we use the formula for integrating exponential functions:

$\int e^{at} \, dt = \frac{e^{at}}{a}$

In this case, $a = -2$, so:

$D(t) = Z \left[ \frac{e^{-2t}}{-2} \right]_0^t$

Now, evaluate the definite integral:

$D(t) = Z \left( \frac{e^{-2t}}{-2} - \frac{e^{0}}{-2} \right)$ $D(t) = Z \left( \frac{e^{-2t}}{-2} + \frac{1}{2} \right)$

Finally, simplify:

$D(t) = \frac{Z}{2} \left( 1 - e^{-2t} \right)$

So, the total distance traveled by the particle during the first $t$ seconds is:

$D(t) = \frac{Z}{2} \left( 1 - e^{-2t} \right) \text{ meters}$

Would you like further clarification or details on any step?

Here are five related questions to deepen understanding:

1. What is the significance of the negative exponent in the velocity function?
2. How does the total distance change as $t$ approaches infinity?
3. What would the result be if the velocity function had a different decay rate, say $e^{-kt}$?
4. How can we calculate the instantaneous acceleration from this velocity function?
5. Can we deduce the initial velocity of the particle from the given velocity function?

Tip: When integrating exponential functions, always pay attention to the coefficient of $t$, as it directly affects the integration process.