Question

a farmer moves along the boundary of a square field of side 20m in the 80s. What will be the magnitude of displacement of the framer at the end of 3 minutes 20 seconds from his initial position​

Answer 1

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A farmer moves along the boundary of a square field of side 20m in the 80s. What will be the magnitude of displacement of the framer at the end of 3 minutes 20 seconds from his initial position

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Length of square field = 20m

Time taken to move along the square field = 80 sec

Total time the farmer walked around the field

= 3 min 20 sec

= 3 × 60 + 20

= 180 + 20

= 200 second

In 80 seconds Farmer covers 80 m

Thus,

Distance covered in 40 seconds = 40 m

Distance covered in 1 second = $\frac{80}{80}$m

Distance covered in 200 second = $\frac{80}{80}$m × 200

= 200 m

Thus,

Total rounds = $\frac{Total\:distance\:covered}{Distance\:covered\:in\:1\:round}$

= $\frac{200}{80}$

= 2.5

∴ Thus farmer takes 2.5 rounds