Question

A farmer moves along a boundary of a square field of aide 10m in 40 sec what qill be the magnitude of displacement of the farmer at the end of 2 min 20 sec

Answer 1

Hey.

Here is the answer.

Perimeter of the square field = 4× side

= 4×10 m = 40 m

Total time given = 2 min 20 s

= 2×60 + 20 s = 120+20 s = 140 s

as he crosses 40m in 40s

so he will cross 140m in 140s

as one round is of 40 m

so he will complete 3 complete round and 20 m more .

As 1 side is 10m ...so 20m is 2 sides.

So he will be at the diagonal point from initial position

so, displacement = √10^2+10^2

= √200 m

= 10√2 m

Thanks.

Answer 2

$\textbf{Solution :}$


Total distance 40m in 40 sec.

Total time taken by the farmer = 2 min 20 sec

=140 seconds. 

Total rounds completed= $\frac{140}{40}$ =3.5 rounds 


In three rounds the displacement is zero.

So, the farmer will be at the diagonal point from staring point.

Displacement = √(10)² + (10)²

= √100 + 100

= √200

= 10√2 m