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?

Asked by Vishal kumar | 1 year ago |  283

##### Solution :-

Given that the farmer covers the entire boundary of the square field in 40 seconds, the total distance travelled by the farmer in 40 seconds is $$4\times(10) = 40\, meters$$

Therefore, the average distance covered by the farmer in one second is: $$\frac{40\,m}{1\,m}=1\,m$$

Two minutes and 20 seconds can be written as 140 seconds. The total distance travelled by the farmer in this timeframe is: $$1\,m\,\times\,140=140\,m$$

Since the farmer is moving along the boundary of the square field, the total number of laps completed by the farmer will be:$$\frac{140}{40}=3.5\,laps$$

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:$$\sqrt{10+10^2}=\sqrt{200}=14.14\,m$$

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 10 m. This is the minimum displacement.

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

Answered by Shivani Kumari | 1 year ago

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