A farmer moves along the boundary of a square field of size 10m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
Farmer takes 40 s to move along the boundary. Thus, after 3.5 round farmer will at point C of the field. Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m north east from initial position.
Answers & Comments
Given:
⟼ Now distance covered by the farmer in 2 min 20 seconds = 1 × 140 = 140m
⟼ Now the total number of rotation the farmer makes to cover a distance of 140 meters =![\frac{Total \:distance}{Perimeter} \frac{Total \:distance}{Perimeter}](https://tex.z-dn.net/?f=%5Cfrac%7BTotal%20%5C%3Adistance%7D%7BPerimeter%7D)
= 3.5
⟼ At this point, the farmer is at a point say B from the origin O
⟼ Thus the displacement s =![\sqrt{ {10}^{2} + {10}^{2} } \sqrt{ {10}^{2} + {10}^{2} }](https://tex.z-dn.net/?f=%5Csqrt%7B%20%7B10%7D%5E%7B2%7D%20%20%2B%20%20%7B10%7D%5E%7B2%7D%20%7D)
⟼ From Pythagoras theorem.
S =![10 \sqrt{2} 10 \sqrt{2}](https://tex.z-dn.net/?f=10%20%5Csqrt%7B2%7D%20)
•°• S = 14.14m
Explanation:
Farmer takes 40 s to move along the boundary. Thus, after 3.5 round farmer will at point C of the field. Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m north east from initial position.