For this weeks of installment of Wasteland Survival with Doctor Albert we show the process of restarting a nuclear power plant after it has gone offline. The energy produced by a nuclear power plant can be used to greatly improve an entire settlement’s quality of life and chances for survival. If you would like to see a walk-through of this process, please view the holotape recording available below.

**Scenario**

In order to repair the plant, there will be times you need to get close to radioactive materials in order to provide repairs to broken infrastructure or to get needed supplies. The example scenario we will consider is that there is something that we need to get that is next to some radioactive waste. We will therefore have to approach the waste grab the item and walk away. During this time, we want to know how much radiation we will be exposed to.

**Distance over time**

Now that we have the scenario described, we will need to determine a few more things. We will begin by finding our distance function. Therefore, let’s list what we have.

- We accelerate constantly from a distance of 5 meters to a distance of 3.
- We then decelerate constantly to a distance of 1 meter.
- Our velocity at 5 meters is 0, at 3 meters is -3 (since distance is getting smaller) and at 1 meter is 0.

On the way to the waster, we therefore have that \[a(t)=\begin{cases} a_{1} \text{ for } 3 \leq s \leq 5 \\ a_{2} \text{ for } 1 \leq s \leq 3 \end{cases}\] where \(a_{1}<0\) (since we are accelerating in a negative direction) and \(a_{2}>0\) are constant. Since velocity is the anti-derivative of acceleration with respect to time, we then have that \[v(t)=\begin{cases} a_{1}t \text{ for } 5 \geq s \geq 3 \\ a_{2}(t-t_{1})-3 \text{ for } 3 \geq s \geq 1\end{cases}\]. Where \(t_{1}\) is the time when we reach 3 meters. Again, noting that position is the anti-derivative of velocity, we have that \[s(t)=\begin{cases} \frac{a_{1}}{2}t^{2}+5 \text{ for } 5 \geq s \geq 3 \\ \frac{a_{2}}{2}(t-t_{1})^{2}-3(t-t_{1})+3 \text{ for } 3 \geq s \geq 1\end{cases}\].

If we knew the time limits for the functions instead of the distance limits, it would make this problem much easier. We aren’t given these, so we will have to find them. In order to do so, we note that from the first case of the position and velocity functions, \[\begin{align*}3&=\frac{a_{1}}{2}\left(t_{1}\right)^{2}+5 \\ -3 & =a_{1}t_{1}. \end{align*}\] We, then have that \[\begin{align*}-\frac{4}{a_{1}}&=t_{1}^{2}=\frac{9}{a_{1}}^{2} \\ a_{1}&=\frac{-9}{4} \\ t_{1}&=\frac{4}{3} \end{align*}\]

If we use a similar process for the second portion of the the scenario, we will find that \(a_{2}=\frac{9}{4}\) and \(t_{2}=\frac{8}{3}\) where \(t_{2}\) is the time we reach a distance of 1 meter. We now have that our position function on the way to the waste is given by \[s(t)=\begin{cases} -\frac{9}{8}t^{2}+5 \text{ for } 0 \leq t \leq \frac{4}{3} \\ \frac{9}{8}(t-\frac{4}{3})^{2}-3(t-\frac{4}{3})+3 \text{ for } \frac{4}{3} \leq t \leq \frac{8}{3}.\end{cases}\]

**Radiation absorbed**

We are now ready to find the total radiation absorbed. In order to do this, we need to find the integral of the rads per second with respect to the time spent moving. We will therefore have that the total radiation absorbed while we move toward the waste, \(r_{t}\), is given by \[\begin{align*} r_{t}&=\int_{0}^{\frac{4}{3}}\frac{12}{(-\frac{9}{8}t^{2}+5)^{2}}dt +\int_{\frac{4}{3}}^{\frac{8}{3}}\frac{12}{(\frac{9}{8}(t-\frac{4}{3})^{2}-3(t-\frac{4}{3})+3)^{2}}dt \\ &=\int_{0}^{\frac{4}{3}}\frac{768}{(9t^{2}-40)^{2}}dt+\int_{\frac{4}{3}}^{\frac{8}{3}}\frac{256}{3(3t^{2}-16t+24)^{2}}dt. \end{align*}\]

In order to complete these integrals, we will have to use partial fraction decomposition on both of them. Then we will be able to find the anti-derivatives using substitution. If you would like to see the step involved in this, just ask and I will send this along to you. Otherwise, we will end up getting \(r_{t}=.91+8.07=8.98\). That is, while running to the get the object you needed, you will absorb about 9 rads.

On the way back, you will follow a symmetric route, so you will absorb the same number of rads again. Therefore, you will absorb a total of 18 rads while going to the pick up the object then return.

**Result**

How much will these 18 rads affect you? In order to determine this, I actually ran this path to see. I’ll let the picture below describe the results.

**Conclusion**

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