# 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 its initial position ?

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by Dhairya

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by Dhairya

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Case 1. Displacement = 10m

Time = 40 sec

Velocity = 10/40 = 0.25 m/sec

Case 2. Velocity = 0.25 m/sec

Time = 2 min 20 sec = 140 sec

Displacement = Velocity * Time

0.25 *140 = 35 m

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Time = 40 sec

Velocity = 10/40 = 0.25 m/sec

Case 2. Velocity = 0.25 m/sec

Time = 2 min 20 sec = 140 sec

Displacement = Velocity * Time

0.25 *140 = 35 m

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4*10m=40m(perimeter of square field )

then after 140 s he will be at opposite corner. or AC(if he starts from A)

by pythagorus theorem

(AC)^2=(AB)^2+(AB)^2

!! = (10M)^2+(10M)^2

!! = 100M^2 + 100M^2

(AC)^2= 200M^2

AC = SQUARE ROOT OF 200M^2

AC = (AFTER PRIME FACTORIZATION OF 200M^2) 10<ROOT>2

he starts from A then in 140 sec he will reach opp. corner

displacement = AC i.e. x

x^2= AB^2 +BC^2

x^2 = 100+100

x^2=200

x=√200

x=14.14m