Question

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

Answer 1

Answer:

14.14m

Explanation:

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.

hope this answer helps you

Answer 2

Answer:

Given that the farmer covers the entire boundary of the square field in 40 seconds, the total distance traveled by the farmer in 40 seconds is 4*(10) = 40 meters.

Therefore, the average distance covered by the farmer in one second is: 40m/40 = 1m

Two minutes and 20 seconds can be written as 140 seconds. The total distance traveled by the farmer in this timeframe is: 1 m * 140 = 140m

Since the farmer is moving along the boundary of the square field, the total number of laps completed by the farmer will be: 140m/40 = 3.5 laps

Now, the total displacement of the farmer depends on the initial position. If the initial position of the farmer is at one corner of the field, the terminal position would be at the opposite corner (since the field is square).

In this case, the total displacement of the farmer will be equal to the length of the diagonal line across the opposite corners of the square.

Applying the Pythagoras theorem, the length of the diagonal can be obtained as follows: √(102+102)= √200= 14.14m.

This is the maximum possible displacement of the farmer.

If the initial position of the farmer is at the mid-point between two adjacent corners of the square, the net displacement of the farmer would be equal to the side of the square, which is 10m. This is the minimum displacement.

If the farmer starts at a random point around the perimeter of the square, his net displacement after traveling 140m will lie between 10m and 14.14m.