Question

A farmer moves along the boundary of a square field of side 10m
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

Given :

• A farmer moves along the boundary of a square of side 10m in 40 seconds.

To Find :

• Magnitude of displacement of the farmer at the end of 2 minutes 20 seconds.

Solution :

ABCD is a square field of the side 10cm . So , the magnitude of the displacement.

$\longmapsto\tt{{AC}^{2}={AB}^{2}+{BC}^{2}}$

$\longmapsto\tt{AC=\sqrt{{AB}^{2}+{BC}^{2}}}$

Here :

• AB = 10m
• BC = 10m

$\longmapsto\tt{AC=\sqrt{{(10)}^{2}+{(10)}^{2}}}$

$\longmapsto\tt{AC=\sqrt{100+100}}$

$\longmapsto\tt{AC=\sqrt{200}}$

$\longmapsto\tt{AC=10\sqrt{2}m.}$

So , The Displacement of the farmer at the end of 2 minutes 20 seconds will be 10√2 m..

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.

Displacement AC=√10^2+10^2

√100+100

√200=10√2

=10×1.414

=14.14m

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