Question

a farmer moves field along the boundary of a square of side 10m in 40s what Will be the magnitude of displacement of the farmer at the end of 2 minutes 20 second​

Answer 1

Given :-

A farmer moves field along the boundary of a square of side 10m in 40s

To Find :-

What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 second​

Solution :-

Side = 10 m

Perimeter = 4 × side

Perimeter = 4 × 10

Perimeter = 40 m

Now,

Speed = Distance/Time

Speed = 40/40

Speed = 1 m/s

2 m 20 sec = 2 × 60 + 20

120 + 20

140 sec

Distance = Speed × Time

Distance = 1 × 140

Distance = 140 m

Total rounds made by farmer in 140 sec

Distance/Perimeter

140/40

14/4

3.5 round

Now

Since, sides of square equal.

Displacement = Diagonal of the square

Diagonal = √2 × Side

Diagonal = √2 × 10

Diagonal = 10√2 m

The displacement is 10√2 m

Answer 2

$\large \bigstar \: \sf \underline\bold{ \underline{Given : }}$

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

$\\ \\ \large \bigstar \: \sf \underline\bold{ \underline{To \: find: }}$

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

$\\ \\ \large \bigstar \: \sf \underline\bold{ \underline{Solution: }} \\ \\</p><p>$

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

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

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

Here :

★ AB = 10m

★ BC = 10m

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

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

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

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

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