Question

A farmer moves along the boundary of a square field of size 10m in 40s. What will be the magnitude of displacement of the farmer at the end of 2 min 20 sec from his initial position??
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Answer 1

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 =

$\sqrt{10m) }^{2} + (10m) ^{2} \\ = \sqrt{100m}^{2} + 100m {}^{2} \\ = 200m {}^{2} \\ = 10 \sqrt{2m}$

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.

Answer 2

Answer:

${\huge{\underline{\bf{\pink{solution}}}}}$

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 =

$\sqrt{10} {m}^{2} + (10m {)}^{2} \\ = \sqrt{100 {m}^{2} } + 100 {m}^{2} \\ = 200 {m}^{2} \\ = 10 \sqrt{2m}$

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.