A farmer moves along the boundary of a square field of side 10m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

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Solution

Step 1: Given data
A farmer moves along the boundary of a square field of side 10m in 40 seconds.
Step 2: Formula used
$S p e e d = \frac{d i s \tan c e}{t i m e}$
Step 3: Calculating the speed of the farmer
We need to find the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position.
The side of square ABCS is $x = 10 m$
The perimeter of the square = $4 \times s i d e = 4 x = 4 \times 10 = 40 m$
It is given that the farmer takes 40 seconds to complete one round of this square field.
Hence the distance travelled by the farmer in 40 seconds is 40m.
Speed of farmer= $\frac{d i s \tan c e}{t i m e} = \frac{40 m}{40 s} = 1 m / s$
Step 4: Calculating distance travelled by the farmer
Now we need to calculate the distance travelled by the farmer in 2 minutes 20 seconds or 140 seconds.
Using the same speed formula, we get
$s p e e d = \frac{d i s \tan c e}{t i m e} 1 = \frac{d i s \tan c e}{140} d i s \tan c e = 140 m$
Hence the distance travelled by the farmer in 2 minutes 20 seconds is 140m.
Step 5: Calculating the magnitude of displacement
Now we need to calculate the position of the farmer on the boundary of square ABCD after 2 minutes 20 seconds.
$P e r i m e t e r o f t h e s q u a r e = o n e r o u n d o f t h e s q u a r e 40 m = 1 r o u n d 1 m = \frac{1}{40 r o u n d} 140 m = \frac{1}{40} \times 140 = 3.5 r o u n d s$
Hence the farmer completes 3.5 rounds of the square field ABCD in 2 minutes 20 seconds.
When the farmer completes one round of square field ABCD starting from vertex A clockwise, after completion the net displacement from the starting position is zero.
Hence the net displacement after the completion of 3 rounds is also zero.
After 0.5 rounds, the farmer reaches the diagonally opposite point of the starting position, that is if the farmer starts from vertex A of square field ABCD, it reaches point C.
The net displacement from the starting point is equal to the diagonal AC of the square field.
Hence the net displacement of the farmer after 3.5 rounds is equal to the diagonal AC of the square field.
Applying Pythagoras theorem in right-angle triangle ADC, we get
$H y p o t e n u s e^{2} = B a s e^{2} + H e i g h t^{2} A C^{2} = A D^{2} + A B^{2} A C^{2} = 10^{2} + 10^{2} A C^{2} = 200 A C = \sqrt{200} A C = 10 \sqrt{2} m$
Hence the net displacement if the farmer after 3.5 rounds is equal to $10 \sqrt{2} m$ .
Therefore, the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position is $10 \sqrt{2} m$ .

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