Question

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

Heya friend,

Side of the square= 10m
Time taken by farmer to complete one round = 40s

[Make a fig of square ABCD, that is the square field]

Displacement = [By using Pythagoras theorem]

DB^2= BC^2+DC^2
DB^2=10^2+10^2
DB^2= 100+100
DB^2= 200
DB= √200
DB= 10√2m
DB=14.1421356

Hope I am correct?

Answer 2

Given, Side of the square field= 10m
Therefore, perimeter = 10 m × 4 = 40 m
Farmer moves along the boundary in 40s.
Displacement after 2 m 20 s = 2 × 60 s + 20 s = 140 s =?
Since in 40 s farmer moves 40 m

Therefore, in 1s 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.

Therefore, Displacement AC = √(10m)²+(10m)²
=√100m²+100m²
=√200m²
= 10√2m
= 10*1.414


From MrMysterious2

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