Q38P - Fundamentals Of Physics

Question

An old manuscript reveals that a landowner in the time of King Arthur held 3.00 acres of plowed land plus a livestock area of 25.0 perches by 4.00 perches. What was the total area in (a) the old unit of roods and (b) the more modern unit of square meters? Here, 1 acre is an area of 40 perches by 4 perches, 1 rood is an area of 40 perches by 1 perch, and 1 perch is the length 16.5 ft.

Step-by-Step Solution

Answer

1. Total area in old units of roods is${}^{\mathbf{14}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{roods}}$.

2. Total area in a modern unit of square meters is${}^{\mathbf{1}\mathbf{.}\mathbf{47}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}{\mathbf{m}}^{\mathbf{2}}}$.

Step 1: Given data

$\mathbf{1}\mathbf{ }\mathbf{acre}\mathbf{ }\mathbf{=}\mathbf{40}\mathbf{ }\mathbf{perch}\mathbf{\times }\mathbf{ }\mathbf{4}\mathbf{ }\mathbf{perch}$

$\mathbf{1}\mathbf{ }\mathbf{rood}\mathbf{ }\mathbf{=}\mathbf{40}\mathbf{ }{\mathbf{perch}}^{\mathbf{2}}$

$\mathbf{1}\mathbf{ }\mathbf{perch}\mathbf{ }\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{ft}$

$\mathbf{1}\mathbf{ }\mathbf{m}\mathbf{ }\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{28}\mathbf{ }\mathbf{ft}$

Step 2: Understanding the unit conversion

Every unit of the area mentioned in the problem has a conversion factor between them and hence the area can be measured in units of rood and square meters.

Step 3: (a) To find a total area in old units

The total area ${}^{\mathbf{A}}$ can be calculated as:

$$\begin{gathered} \mathbf{A}\mathbf{=}\mathbf{3}\mathbf{ }\mathbf{acres}\mathbf{ }\mathbf{+}\left(25 \mathrm{perch}\times 4 \mathrm{perch}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{3}\mathbf{ }\mathbf{acres}\mathbf{+}\mathbf{ }\mathbf{100}\mathbf{ }{\mathbf{perch}}^{\mathbf{2}} \end{gathered}$$

Now, you need to convert the acres into ${}^{{\mathbf{perch}}^{\mathbf{2}}}$ .

$$\begin{gathered} \mathbf{A}\mathbf{ }\mathbf{=}\left(3 \mathrm{acres}\times \frac{40 \mathrm{perch} \times 4 \mathrm{perch}}{1 \mathrm{acres}}\right)\mathbf{ }\mathbf{+}\mathbf{100}\mathbf{ }{\mathbf{perch}}^{\mathbf{2}}\mathbf{)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{480}\mathbf{ }{\mathbf{perch}}^{\mathbf{2}\mathbf{ }}\mathbf{+}\mathbf{ }\mathbf{100}\mathbf{ }{\mathbf{perch}}^{\mathbf{2}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{580}\mathbf{ }{\mathbf{perch}}^{\mathbf{2}} \end{gathered}$$

Now, you need to convert this area into roods. By using a conversion factor,

$\mathbf{1}\mathbf{ }\mathbf{rood}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{40}\mathbf{ }{\mathbf{perch}}^{\mathbf{2}}$

$\mathbf{A}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{580}\mathbf{ }{\mathbf{perch}}^{\mathbf{2}}\mathbf{ }\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{rood}}{\mathbf{40}\mathbf{ }{\mathbf{perch}}^{\mathbf{2}}}$

Thus, total area is ${}^{\mathbf{14}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{roods}}$.

Step 4: (b) Determination of the total area in modern units

It is given that, ${}^{\mathbf{1}\mathbf{ }\mathbf{perch}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{16}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{ft}}$

Therefore,

${}^{\mathbf{580}\mathbf{ }\mathbf{p}\mathbf{e}\mathbf{r}\mathbf{c}{\mathbf{h}}^{\mathbf{2}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{1}\mathbf{,}\mathbf{58}\mathbf{,}\mathbf{177}\mathbf{.}\mathbf{25}\mathbf{ }\mathbf{f}{\mathbf{t}}^{\mathbf{2}}}$

You can say that total area is ${}^{\mathbf{1}\mathbf{.}\mathbf{58}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{ }{\mathbf{ft}}^{\mathbf{2}}}$ .

Therefore, area in modern units is,

$$\begin{gathered} \mathbf{A}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{1}\mathbf{.}\mathbf{58}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}\mathbf{ }{\mathbf{ft}}^{\mathbf{2}}}\mathbf{\times }{\left(\frac{1 m}{3.28 \mathrm{ft}}\right)}^{\mathbf{2}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{47}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{ }{\mathbf{m}}^{\mathbf{2}} \end{gathered}$$

Thus, the total area in modern units of square meters is ${}^{\mathbf{1}\mathbf{.}\mathbf{47}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{ }{\mathbf{m}}^{\mathbf{2}}}$ .