Question

Heya

A 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 1

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.

$Therefore, \: Displacement \: AC = \sqrt{ {(10m)}^{2} + ( {10m)}^{2} } \\ = \sqrt{ {100m}^{2} + {100m}^{2} } \\ = \sqrt{ {200m}^{2} } \\ = 10 \sqrt{2} m \\ = 10 \times 1.414 \\</p><p></p><h2> = 14.14m$

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

$Hope \: This \: Will \: help \: you$❤

Answer 2

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,DisplacementAC=

(10m) 2 +(10m) 2

= 100m 2 +100m 2

=

200m

2

=10

2

m

=10×1.414

=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 it Helps You ❣️