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

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The farmer takes 40 s to cover 4 × 10 = 40 m. In 2 min and 20 s (140 s), he will cover a distance Therefore, the farmer completes rounds (3 complete rounds and a half round) of the field in 2 min and 20 s. That means, after 2 min 20 s, the farmer will be at the opposite end of the starting point. Now,...
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The farmer takes 40 s to cover 4 × 10 = 40 m.

In 2 min and 20 s (140 s), he will cover a distance

Therefore, the farmer completes rounds (3 complete rounds and a half round) of the field in 2 min and 20 s.

That means, after 2 min 20 s, the farmer will be at the opposite end of the starting point.

Now, there can be two extreme cases.

Case I: Starting point is a corner point of the field.

In this case, the farmer will be at the diagonally opposite corner of the field after 2 min 20 s.

Therefore, the displacement will be equal to the diagonal of the field.

Hence, the displacement will be

Case II: Starting point is the middle point of any side of the field.

In this case the farmer will be at the middle point of the opposite side of the field after 2 min 20 s.

Therefore, the displacement will be equal to the side of the field, i.e., 10 m.
 

For any other starting point, the displacement will be between 14.1 m and 10 m.

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Total time is 140s Time required to complete one complete moment along border is 40s So we can cover 3 and 1/2 moment along border. At last we are at point of hypotenuse so displacement is length of hypotenuse By that it is 10√2 m
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Total  time  is 140s 

Time required to complete one complete moment  along  border is 40s 

So we can cover 3 and 1/2 moment along border. 

At last we are at point of hypotenuse so displacement  is length of hypotenuse 

By that it is 10√2 m

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1 Comments

Completed graduation in Mathematics from Calcutta University

The displacement of the man will be 14.14 m
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The displacement is the diagonal of a squareAnsA= 10√2
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