Question

A farmer moves along the boundary of the square field of side $10\text{m}$ in $40\text{seconds}$. What will be the magnitude of displacement of the farmer at the end of $2\text{minutes}20\text{seconds}$ from its initial position if he starts from one of the corners.

• A. $0\text{m}$
• B. $14.1\text{m}$
• C. $30\text{m}$
• D. $20\text{m}$

Answer 1

The correct option is B $14.1\text{m}$

Let the initial point of the farmer be point $\text{P}$.

The distance covered by the farmer in $40\text{s}$ is $4\times 10=40\text{m}$.

The total time given: $2\text{min}$ and $20\text{s}=(2\times 60)+20=140\text{s}$

Since the farmer moves $40\text{m}$ in $40\text{s}$, therefore, in $1\text{s}$ the farmer covers a distance,
$\frac{40}{40}\text{m}=1\text{m}$

Therefore, the distance covered by the farmer in $140\text{s}$ is:

$140\times 1\text{m}=140\text{m}$.

Thus the number of rotations to cover
$140\text{m}$ along the boundary is:

$\frac{\text{Distance}}{\text{Perimeter}}=\frac{140}{40}=3.5\text{rotations}$

So, the farmer will be at point $R$ after $140\text{s}$.

So the magnitude of displacement $=PR=\sqrt{{10}^2+{10}^2}=14.1\text{m}$