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 secondsAccording 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

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 90^{o} angle between them

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

(AC)^{2} = (AB)^{2} + (BC)^{2}

(AC)^{2} = (10)^{2} + (10)^{2 }(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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