Question

UIC SCHO
4. A farmer moves along the boundary of a square field of side 10
what will be the magnitude of displacement of the farmer at the
minutes and 20 sec?
-zian with a​

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.

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:

Given side of square =10m, thus perimeter P=40m

Time taken to cover the boundary of 40 m =40 s

Thus in 1 second, the farmer covers a distance of 1 m

Now distance covered by the farmer in 2 min 20 seconds = 1×140=140m.

Now the total number of rotation the farmer makes to cover a distance of 140m

meters = total distance / Perimeter.

= 3.5

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

Thus the displacement s = √10² +10 ²

From Pythagoras theorem.

s=10 √2

=14.14m