Question

2. 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 1

Answer:

Given side of square =10m, thus perimeter P=40m

Time taken to cover the boundary of 40 m =40 s

Thus in 1 second, the farmer covers a distance of 1 m

Now distance covered by the farmer in 2 min 20 seconds = 1×140=140m

Now the total number of rotation the farmer makes to cover a distance of 140 meters =  

    total  distance

    --------------------

    perimeter

​  

=3.5  

At this point, the farmer is at a point say B from the origin O

Thus the displacement s=  

-/10sq.+10sq. from Pythagoras theorem.

 

s=10  -/2

​  

=14.14m

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Answer 2

Answer:

Here, Side of the given square field = 10m

so, perimeter of a square = 4*side = 10 m * 4 = 40 m

Farmer takes 40 s to move along the boundary.

Displacement after 2 minutes 20 s = 2 * 60 s + 20 s = 140 seconds

since in 40 s farmer moves 40 m

Therefore, in 1s the distance covered by farmer = 40 / 40 m = 1m

Therefore, in 140s distance covered by farmer = 1 � 140 m = 140 m. Now, number of rotation to cover 140 along the boundary= Total Distance / Perimeter

= 140 m / 40 m = 3.5 round

Thus, after 3.5 round farmer will at point C of the field.

Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m north east from initial position.

Explanation:

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