A farmer moves along the boundary of a square field of side 10m in 40s. What will be the magnitude of displacement of the farmer at the end of 2 minutes and 20 seconds?
Answers
Answer:
After 3.5 round farmer will at point C of the field.
Explanation:
Here,
Side of the given square field = 10m
So,
perimeter of a square = 4*side = 10 m × 4 = 40 m
Farmer takes 40 s to move along the boundary.
Displacement after 2 minutes 20 s = 2 × 60 s + 20 s = 140 seconds
Since in 40 s farmer moves 40 m
- In 1s the distance covered by farmer = 40 / 40 m = 1m
- In 140s distance covered by farmer = 1 × 140 m = 140 m.
Now,
Number of rotation to cover 140 along the boundary =
= 140 m / 40 m = 3.5 round
Thus,
After 3.5 round farmer will at point C of the field.
- Displacement of farmer at the end of 2 minutes and 20 seconds
- Side of square field = 10m
- farmer moves along the boundary of a square field of side 10m in 40s.
we know that,
☘️ perimeter of square = 4 × side
⪼ perimeter of square = 4 × 10
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀= 40m
Time taken by farmer to cover 40m = 40s
So,
In 1 second the farmer covers 1meter distance.
Now,
Distance covered by the farmer at the end of 2 minutes and 20 seconds:-
Time = 2minute 20 second
Converting 2 minutes into seconds:-
= 2 × 60 + 20
= 120 + 20
= 140 second
As we know that, the farmer covers 1m distance in 1 sec ,
So,
Distance covered by the farmer at the end of 2 minutes and 20 seconds is :-
= 1 × 140
= 140m
So,
The farmer will cover 140m distance in 2 minutes and 20 seconds.
Now,
Number of rotation the farmer takes to cover the distance
= Total Distance/perimeter
= 140/40
= 3.5
If the farmer stars from A( initial position) then after 3.5 rounds he was at point C.
Therefore, Displacement is AC.
● By using Pythagoras theorem,
- AC² = AB² + BC²
➛ AC² = 10² + 10²
➛ AC² = 100 + 100
➛ AC² = 200
➛ AC = √200
➛ AC = 10√2
- ● AC = 14.14m
Hence Displacement of farmer at the end of 2 minutes and 20 seconds is 14.14m
━━━━━━━━━━━━━━━━━━━━━━━━━