- 2. A farmer moves along theboundary of a square field of side10 m in

1.	- 2. A farmer moves along theboundary of a square field of side10 m in 40 s. What will be themagnitude of displacement of thefarmer at the end of 2 minutes 20seconds from his initial position?
Answer» Side of square = 40 m Perimeter of Square = 4side= 440 = 160 m In 40 sec farmer complete one round of boundary it means farmer cover distance = 160 min 40 sec but displacement is 0as starting and end point is same. Now,2 min 20 sec = 260 + 20= 120 + 20 = 140 Therefore,In 140 sec farmer cover distance = 160140/40= 4140= 560 m or 3.5 rounds of square boundary In 3 rounds displacement is 0For half round displacement is equal to diagonal of square = root(2)side= 40root(2)= 401.414= 56.56 m

- 2. A farmer moves along theboundary of a square field of side10 m in 40 s. What will be themagnitude of displacement of thefarmer at the end of 2 minutes 20seconds from his initial position?

Answer»

Side of square = 40 m

Perimeter of Square = 4*side= 4*40 = 160 m

In 40 sec farmer complete one round of boundary it means farmer cover distance = 160 min 40 sec but displacement is 0as starting and end point is same.

Now,2 min 20 sec = 2*60 + 20= 120 + 20 = 140

Therefore,In 140 sec farmer cover distance = 160*140/40= 4*140= 560 m or 3.5 rounds of square boundary

In 3 rounds displacement is 0For half round displacement is equal to diagonal of square = root(2)*side= 40*root(2)= 40*1.414= 56.56 m

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