Question

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 1

Answer:10 root2 m

Explanation:

SIDE OF THE SQUARE = 10 M

PERIMETER OF THE SQUARE = 4×10 =40 M

HE COMPLETES 1 ROUND IN 40 SECONDS.

SO, SPEED = 40/40= 1 M/S

SO, DISTANCE COVERED IN 2 MIN 20 SECOND OR 140 SECONDS IS = 140 × 1= 140 M

NUMBER OF ROUNDS OF THE SQUARE COMPLETED IN MOVING THROUGH 140 M IS = 140/40 = 3.5

IN 3 ROUNDS THE DISPLACEMENT IS 0. IN 0.5 ROUND THE FARMER REACHES THE DIAGONALLY OPPOSITE END OF THE SQUARE FROM HIS STARTING POINT.

DISPLACEMENT = AC = root(AB^2 + BC^2)

= root(100+100)

= root 200

= 10 root2 m

HOPE IT HELPS YOU!

Answer 2

Answer:

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