Question

A farmer moves along the boundary of a square field of side 10m in 40s what will be the magnitude of
displacement of the farmer at the end of 2 minutes and 20 seconds from his initial position

Answer 1

Side of the square = 10 m

Perimeter of the square = 4×10 = 40 m

He completes 1 round in 40 s.

So, speed = 40/40 = 1 m/s

So, distance covered in 2 min 20 s or 140 s is = 140 × 1 = 140 m

Number of rounds of the square completed in moving through 140 m is = 140/40 = 3.5

In 3 rounds the displacement is zero. In 0.5 round the farmer reaches the diagonally opposite end of the square from his starting point.

Answer 2

Answer:

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.

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