Given Data:
The rat travels 10 in/sec when 2 grams is attached.
Since the distance tarved and the weight are in inversely propotional, it can be written as,
![d=(k)/(W)](https://img.qammunity.org/2023/formulas/mathematics/college/3zt4fgf4hsm63w5fgq9qwndkzdcam1usz8.png)
Here, k is a proportionality constant, and W is the weight added.
When 2 gram added the rat traveled 10 in/sec.
![\begin{gathered} d=(k)/(W) \\ 10=(k)/(2) \\ k=10*2 \\ k=20 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zkf5kqb0rid87gsfjyj3kb8sn3jychliyk.png)
If the rate traveled 18 in/sec the weight added is,
![\begin{gathered} d=(k)/(W) \\ 18=(k)/(W) \\ W=(k)/(18) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wsojet31lopipffghjl7obwg4h5swnpc82.png)
Substitute value of k in the above expression.
![\begin{gathered} W=(20)/(18) \\ =(10)/(9) \\ =1.11\text{ gram.} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ruu6bqoyijfsho70w6lo5k4o42werzf77r.png)
Thus, the rat travels 18 in/sec when added 1.1 grams of weight.