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CBSE Class 9 Answered

A farmer moves along the boundary of square field of side 10m in 40sec . What will be the magnitude of displacement of farmer at the end of 2 min 20 sec from his initial position?
Asked by kanikathakkar836 | 19 May, 2020, 11:27: PM
answered-by-expert Expert Answer

Let the initial point of the farmer be point P

The distance covered by the farmer in 40 s is 4 × 10 = 40 m. 

The total time given: 2 min and 20 s = (2 × 60) + 20 = 140 s

Since the farmer moves 40 m in 40 s, therefore in 1 s the farmer covers a distance,
begin mathsize 12px style 40 over 40 straight m space equals space 1 space straight m end style

Therefore, the distance covered by the farmer in 140 s is:

140 × 1 m = 140 m.

Thus the number of rotations to cover 140 m along the boundary is:

begin mathsize 12px style Distance over Perimeter space equals space 140 over 40 space equals space 3.5 space rotations end style

So, the farmer will be at point R after 140 s.

So the displacement = PR = begin mathsize 12px style square root of 10 squared space plus space 10 squared end root end style = 14.1 m

Answered by Shiwani Sawant | 20 May, 2020, 12:07: AM

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