Q. 93.8( 43 Votes )

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 from his initial position?


Answer :

First we will convert the time into seconds

Total time = 2 minutes 20 seconds

= 2× 60 seconds + 20 seconds

= 120 seconds + 20 seconds

= 140 seconds

 According to Question

 If in 40 seconds farmer makes⇒ 1 Round
 
Then in 140 Seconds, The Farmer makes⇒ 1/40 x 140 Rounds
                                                   = 3.5 Rounds

If the farmer starts from position A (see figure), then after three complete rounds, he will reach at starting position A.



But in the next half round, the farmer will move from A to B, and B to C, so that his final position will be at C.

Hence, the total displacement(Shortest Distance) of the farmer will be AC.


Now,We know ABC is a right angled triangle as all the sides are of equal length with 90o angle between them


So, Using Pythagoras Theorem,  (hypotenuse)2=(base)2+(perpendicular)3

(AC)2 = (AB)2 + (BC)2

(AC)2 = (10)2 + (10)(Given in the Question)

(AC)2 = 100 + 100

(AC)2 = 200

AC = 14.143 m

Hence, the magnitude of displacement( Distance from A To C) of the farmer at the end of 2 minutes and 20 seconds will be 14.143 meters.

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