Question

A farmer moves along the boundary of a square field of side 10 m in 40 sec What will be magnitude of displacement  of the farmer at the end of 2 minutes 20 seconds from his initial position?
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Answer 1

Answer:

side of square field a = 10m

time = 2min + 20sec

distance covered in 40sec = 4×10m = 40m

,, ,, ,, ,,, in 140sec = (40/40)×140

= 140m

displacement = shortest distance between initial and final points

140m = 3×(boundary of field) +20m

so, displacement = √10²+10²

= √ 200

= 10√2m answer

Answer 2

Answer:

14.14m

Explanation:

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 =perimetertotaldistance

=3.5 

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

Thus the displacement s=102+102 from Pythagoras theorem.

s=102=14.14m