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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?
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Solution
Given side of square field = 10m
Then the perimeter of that square field = 4 × 10
= 40m
Total time = 2 minute 20 sec
= (2 × 60 + 20)
= 140 sec
Hence, the distance covered in going around the square field = perimeter of the field
Total distance covered in 40 sec = 40m
Total distance covered in 140 sec = `(40/40) xx140` m
= 140 m
This number of rounds = `140/40`
= 3.5
In 140 sec, that farmer will take 3.5 rounds.
Displacement = Diagonal of square = `(102+102) 1/2`
= `(200)1/2`
= `10sqrt2` m
Hence, magnitude of displacement = `10sqrt2` m
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Fill in the boxes.
| S.No | First Move | Seconde Move | Distance (m) | Displacement |
| 1. | Move 4 metres east | Move 2 metres west | 6 | 2 m east |
| 2. | Move 4 metres north | Move 2 metres south | ||
| 3. | Move 2 metres east | Move 4 metres west | ||
| 4. | Move 5 metres east | Move 5 metres west | ||
| 5. | Move 5 metres south | Move 2 metres north | ||
| 6. | Move 10 metres west | Move 3 metres east |
Write a note on the following:
Nautical mile
The displacement covered by the second hand of radius V in a clock after one revolution is ______.
If the displacement-time graph of a particle is parallel to the time axis, the velocity of the particle is ______.
