1.

HeyaA farmer moves along the boundary of a square field of side 10 m in 40s.what will be the magnitude of displacement of the farmer at the end of 2 minute and 20 seconds ?​

Answer»

ANSWER:

Here, Side of the given square field = 10m

so, perimeter of a square = 4*side = 10 m * 4 = 40 m

Farmer takes 40 s to move ALONG the boundary.

Displacement after 2 minutes 20 s = 2 * 60 s + 20 s = 140 seconds

since in 40 s farmer moves 40 m

Therefore, in 1S the distance COVERED by farmer = 40 / 40 m = 1m

Therefore, in 140s distance covered by farmer = 1 � 140 m = 140 m.

Now, number of rotation to cover 140 along the boundary= Total Distance / Perimeter

= 140 m / 40 m = 3.5 ROUND

Thus, after 3.5 round farmer will at point C of the field.

Therefore, \: Displacement \:  AC =  \sqrt{ {(10m)}^{2}  + ( {10m)}^{2} }  \\  =  \sqrt{ {100m}^{2} +  {100m}^{2}  } \\  =  \sqrt{ {200m}^{2} }  \\  = 10 \sqrt{2} m \\  = 10 \times 1.414 \\  = 14.14m

Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m north east from initial position.

Hope \:  This  \: Will  \: help  \: you



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