CBSE Notes, Lectures

CBSE - Physics - Motion

Motion

NCERT In Text

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?


Given, Side of the square field= 10m
Therefore, perimeter = 10 m x 4 = 40 m
Farmer moves along the boundary in 40s.
Displacement after 2 m 20 s = 2 x 60 s + 20 s = 140 s =?
Since in 40 s farmer moves 40 m

Therefore, in 1s distance covered by farmer = 40 / 40 m = 1m

Therefore, in 140s distance covered by farmer = 1 X 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.

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