Answer

• Magnitude of DISPLACEMENT of the farmer at end of 2 metre 20 seconds from its initial position is 14.14 m

Given

• A farmer moves along the boundary of a square field of side 10 metre in 40 seconds

To Find

• The magnitude of displacement of the farmer at end of 2 min 20 seconds from its initial position​

Concept Used

Displacement : It is defined as the shortest DISTANCE b/w two points , i.e., final and initial points .

Solution

Side of the square field , s = 10 m

⇒ Perimeter , P = 4s

⇒ P = 4 × 10

⇒ P = 40 m

So farmer moves through the boundary of square field in 40 s .

⇒ 40 m ⇔ 40 s

⇒ 1 m = 1 s

Here , Farmer takes 2 min and 20 s totally to move .

⇒ Total time , t = 2 min + 20 s

⇒ t = 2 ( 60 ) s + 20 s

⇒ t = 120 s + 20 s

⇒ t = 140 s

Since farmer moves a meter per second .

So , distance covered in 140 s = 140 m

Now , No. of rotations completed by Farmer = Total distance / Perimeter

⇒ 140 / 40

⇒ 14 / 4

⇒ 3.5 rounds

For 3 complete rounds displacement is zero . Since closed path .

From attachment ,

For half ROUND STARTING from A Farmer will be at C .

So , apply Pythagoras theorem for finding displacement , d.

⇒ d² = 10² + 10²

⇒ d² = 100 + 100

⇒ d² = 200

⇒ d = 10√2 m

⇒ d = 14.14 m

So , Magnitude of displacement off the farmer at end of 2 metre 20 seconds from its initial position is 14.14 m