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?

Answer 1

Answer:

Displacement of the farmer in 2 minutes and 20 seconds = 10√2 meters

Step by step explanations :

given that,

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

here,

side of the square feild = 10 m

so,

its boundary length = its perimeter

perimeter of the square = 4 × side

= 4 × 10 = 40 m

so,

perimeter of the square = 40 m

According to the question,

Farmer moves 40 meter in 40 seconds

given that,

farmer continued moving on the boundary of the square field

for 2 minutes 20 seconds

120 + 20

= 140 seconds

now,

number of revolution of square field

140 = 40 + 40 + 40 + 20

so,

here is 3 complete revolution

and 20 m extra

and

in, 20 meter he will go one corner to the another

so,

displacement = diagonal of the square field

diagonal of square = √2 a

where,

a is the side of the square

so,

length of diagonal = 10√2 m

so,

Displacement of the farmer in 2 minutes and 20 seconds

= 10√2 meters

Answer 2

Answer:

displacement of the farmer = 10√2 m

Explanation:

given the length of side the square feild

= 10 m

A farmer moves along the boundry in 40 s

perimeter of the feild = 4 × side

4 × 10

perimeter = 40 m

now given

farmer moves for 2 minutes 20 seconds

= 140 seconds

he will move around boundary =

140

= 120 + 20

(40 ×3) + 20

= 3 rounds

and moves more for 20 seconds

distance moved in 20 seconds

= 40/2 m

= 20 m

= two adjescent sides of the feild

so,

displacement = diagonal of the square feild

diagonal = √(10² + 10²)

= √(100 + 100)

= √200

= 10√2 m

so,

displacement of the farmer = 10√2 m