Question

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A farmer moves along the boundary of a square field of side 10 m 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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Answer 1

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Answer 2

$\huge\underbrace\purple{\dag Answeř\dag}$

☞☞ Given side of square =10 meters

• Thus, perimeter of square= 4 × sides

= 4 × 10 meters

= 40 meters

↬Time taken to cover the boundary of 40 meters = 40 seconds..

↬Thus, in 1 second, the farmer

covers a distance of 1 meter..

↬↬Now distance covered by the farmer in 2 minutes 20 seconds = 1 × 140 meters

= 140 meters

↬↬Now the total number

of rotation the farmer

makes to cover a distance

of 140 meters = ⤵️⤵️

$( \frac{Total \: \: Distance}{Perimeter})= ) \: remember \: this \: formulae$

Total Distance = 140 meters

Perimeter = 40 meters

↬ 140/40

↬ 3.5 meters

• At this point, the farmer is at a point say B from the origin "O"

• Thus the displacement ⤵️⤵️

$s \: = \: \sqrt{ {10}^{2} + {10}^{2} } \\ from \: \: Pythagoras \: \: theorem.$

$\large\sf \bf \underline\purple{s \: = \: 10 \sqrt{2 } \: meters \\ \: \: = 14.14 \: meters}$