Question

A farmers moves along the boundary of a squase 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 1

Answer:

→ Hey Mate,

→ Given Question : A farmers 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?​

Explanation:

→  At the end of 2 minutes  20 seconds , the farmer will be at c .

→ Displacement = AC

→  AC² = AB ² + BC²

→         =  10²  + 10²

→         =   100 + 100

→         =  200

→ AB = 300 m

→ AC= 100 m

→ consider an object starting from A and moving along a straight line 300 m to B It then reverses the direction and moves back 200 m to c.

→ Distance AC = AB + BC

→                       = 300 + 200

→               AC        =  500 m

→ Displacement AC = 100 m

→ Distance and Displacement will be equal if an object is traveling in a straight line without reversing its direction.

→ At the end of 2 min 20 sec , the farmer will be at c .

→ Displacement = AC

→ AC² = AB² + BC²

→        = 10² + 10²

→        =  100 + 100

→        =  200

→        =  100 × 2

→  AC      =  √ 100 × 2

→             =  10√ 2

→ THANK YOU

Answer 2

Farmer takes 40 s to move along the boundary. 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.

Given the problem, a farmer moves along the boundary of a square field of side 10m in 40 seconds.

We need to find the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position.

First, we calculate the perimeter of the square.

Hence the distance travelled by the farmer in 2 minutes 20 seconds is 140m.

Now we need to calculate the position of the farmer on the boundary of square $ABCD$ after 2 minutes 20 seconds.