Question

A farmer moves along the boundary of a square field of side 10 metre in 40 seconds what will be the magnitude of displacement of the farmer at the end of 2 minutes 20 second from his initial position

Answer 1

Answer:

Given, 1round=40sec

time taken to travel =2min20sec=140sec

40sec=40min

1sec=40/40=1m

140sec=(1*140)=140m

Thus, distance =140m

displacement =AC

By Pythagoras theorem, we have

AC^2=AB^2+BC^2

=10^2+10^2

=200

AC=√200=10√2

Hence, displacement =10√2m

Answer 2

Answer:

$\boxed{\bold{10\sqrt{2}}}$

Explanation:

Boundary of square field =10m

Perimeter = 4*side = 40 m

time taken = 40 seconds

hence,

speed = distance /time

speed of farmer = 40/40 = 1 m/sec

$\rule{200}{2}$

Time given = 2 min 20 sec = 140 seconds

distance covered =  speed *time

Distance covered = 140 m

now,

Number of Rounds =  140/40 = 2.5 rounds

$\rule{200}{2}$

Thus,

The farmer will be on the opposite corner of the square field from the point he started.

Magnitude of the displacement will be along the diagonal of the square field.

so,

Displacement=

$=\sqrt{10^2+10^2}\\ \\ \\ \\=\sqrt{200}\\ \\ \\ \\=10\sqrt{2} m$