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? In physics

Answer 1

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

Side of square = 10 m and time = 40 sec

Perimeter of square = area × area

= 4 × 10 = 40 m

We have to find the displacement of the farmer at the end of 2 minutes 20 seconds.

Time = 2 min 20 sec

1 min = 60 sec

2 min = 2(60) = 120 sec

= 120 sec + 20 sec = 140 sec

Now,

In 1 sec distance covered by farmer = 40/40 = 1 m

So, in 140 sec distance covered by farmer = 1 × 140 = 140 m

Number of rotations to cover 140 m along the boundary = Distance/Perimeter

= 140/40 = 3.5 rounds

Therefore, the farmer takes 3.5 revolutions.

Let us assume that farmer is at the point A from the origin of the square field.

Now,

Displacement = diagonal of square

And from above we have a side of square = 10 m

So, displacement = 10√2 m.

MARK ME AS BRILLIANT.

Answer 2

10√2m.............

step by step explanation:

farmer moves in square Field

therefore, displacement = daigonal of field

we know that , daigonal of square = side x √2

= 10√2m