Question

a farmer move along the boundary of a square field of side 10 m in 40 s. what will be the magnitude of displacement of the farms at the end of 2 min 20 sec from his initial position ? ​

Answer 1

Answer:

2.  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 from his initial position?

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.

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

Explanation:

Answer 2

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

Side of square = 10 m and time = 40 sec

Perimeter of square = 4 × side

= 4 × 10 = 40 m

We have to find the displacement of the farmer at the end of 2 min 20 sec.

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

Therefore, the displacement of the farmer is 10√2 m.