Rishi is a farmer who owns a small piece of farmland in the form of a rectangle with length and breadth of 40.25 m and 30.75 m respectively. Rishi moves along the boundary of her farmland starting from one corner (assumed to be origin) and covers half of the boundary of her farmland.
(i) Calculate the area of Rishi’s farmland up to correct significant figures in both m² as well as in cm²
Area Calculation:
The area A of Rishi’s rectangular farmland is calculated using the formula for the area of a rectangle:
A = \text{length} \times \text{breadth}
Given the length is 40.25 m and the breadth is 30.75 m, the area in square meters is:
A = 40.25 \, \text{m} \times 30.75 \, \text{m} = 1237.6875 \, \text{m}^2
Significant Figures in m²:
To express the area to the correct number of significant figures, we take into account the measurements provided. Both measurements are given to four significant figures. Thus, we will round the area to four significant figures:
A \approx 1237.69 \, \text{m}^2
Conversion to cm²:
Since 1 m = 100 cm, 1 m² = 10,000 cm². So, to convert the area from m² to cm², we multiply by 10,000:
A = 1237.69 \, \text{m}^2 \times 10,000 \, \frac{\text{cm}^2}{\text{m}^2} = 12376869 \, \text{cm}^2
Rounded to the appropriate significant figures:
A \approx 12376900 \, \text{cm}^2
Let’s summarize both calculations:
- Area in square meters (m²): 1237.69 \, \text{m}^2
- Area in square centimeters (cm²): 12376900 \, \text{cm}^2
(ii) Express the net displacement vector of Rishi in terms of unit vectors along coordinate axes by drawing the relevant diagram.
Displacement Vector:
Rishi starts at the origin and moves half the perimeter of the rectangle. Let’s calculate the perimeter first:
Perimeter P of the rectangle is:
P = 2 \times (\text{length} + \text{breadth}) = 2 \times (40.25 \, \text{m} + 30.75 \, \text{m}) = 142 \, \text{m}
Since Rishi covers half of this perimeter:
\frac{P}{2} = \frac{142 \, \text{m}}{2} = 71 \, \text{m}
Rishi begins at the origin (0, 0) and follows the boundary: first moving 40.25 m along the x-axis, then 30.75 m along the y-axis.
For half the perimeter, Rishi ends at:
- 40.25 \, \text{m} along the x-axis.
- 30.75 \, \text{m} along the y-axis.
Thus, Rishi does not complete the next 40.25 m fully along the negative x-axis but rather ends at point (40.25 - y_{remaining}, 30.75).
Since she travels a total of 71 m, we need to find the remaining 71 - (40.25 + 30.75) = 0 \, \text{m}, meaning she stops right at the y-end.
Thus, the net displacement vector \vec{d} in terms of unit vectors \hat{i} and \hat{j} (for x and y directions, respectively) is:
[
\vec{d} = 40.25 , \hat{i} + 30.75 , \hat{j}
]
Diagram Representation:
Diagram not possible in text format, but we would typically illustrate a rectangle, showcasing the path from the origin (bottom left corner) to the top right corner tracing the path.
(iii) State the law of vector addition which you used to find Rishi’s net displacement.
To determine Rishi’s net displacement, we use the Parallelogram Law of Vector Addition, which states that if two vectors are represented as adjacent sides of a parallelogram, then the resultant vector (here, displacement) is represented by the diagonal of the parallelogram that starts from the same point.
In Rishi’s case, her horizontal and vertical displacements are thought of as sides of a parallelogram or simply added head to tail to result in the net displacement vector.
(iv) Find the dot and cross product of Rishi’s horizontal and vertical displacement.
Dot Product:
Given vectors \vec{x} = 40.25 \, \hat{i} and \vec{y} = 30.75 \, \hat{j}, the dot product is:
\vec{x} \cdot \vec{y} = (40.25 \, \hat{i}) \cdot (30.75 \, \hat{j}) = 40.25 \times 30.75 \times (\hat{i} \cdot \hat{j}) = 40.25 \times 30.75 \times 0 = 0
The dot product is zero because \hat{i} and \hat{j} are orthogonal vectors (perpendicular to each other).
Cross Product:
The cross product is given by:
\vec{x} \times \vec{y} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 40.25 & 0 & 0 \\ 0 & 30.75 & 0 \end{vmatrix} = (40.25 \times 0 - 0 \times 30.75) \hat{k} = 40.25 \times 30.75 \hat{k}
Thus, the magnitude of the cross product is:
|\vec{x} \times \vec{y}| = 40.25 \times 30.75 = 1237.6875 \, \hat{k}
However, as we’re looking for significant figures:
|\vec{x} \times \vec{y}| \approx 1237.69 \, \hat{k}
The direction of the cross product is perpendicular to the plane formed by the vectors, along the z-axis in the \hat{k} direction.
In summary, for Rishi’s boundary path:
-
Area of Farmland:
- 1237.69 \, \text{m}^2
- 12376900 \, \text{cm}^2
-
Net Displacement Vector:
[ \vec{d} = 40.25 , \hat{i} + 30.75 , \hat{j} ]
-
Law Used:
- Parallelogram Law of Vector Addition.
-
Products:
- Dot Product: 0
- Cross Product: [ 1237.69 , \hat{k} ]
@anonymous6, I hope this helps clarify the details regarding Rishi’s farmland and movements!