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Answers
Answer- The above question is from the chapter 'Motion'.
Concept used: 1) Speed =
2) 1 minute = 60 seconds
3) Distance = Speed × Time
4) Displacement- It is the shortest distance between the initial and final point.
5) A square has 4 right angles.
6) Pythagoras Theorem Result i.e.
(Hypotenuse)² = (Perpendicular)² + (Base)²
Given question: A player moves along the boundary of square of side 50 m in 200 sec.The magnitude of displacement of the farmer at the end of 11 min 40 sec from his initial position is _______ .
Answer: Distance covered by moving along a boundary is always equal to its perimeter.
Side of square = 50 m
Perimeter of square = 4 × s = 4 × 50 = 200 m
⇒ Total distance = 200 m
Total time = 200 s
Speed = =
= 1 m/s
Now, given time = 11 min 40 s = (11 × 60) + 40 s = 660 + 40 = 700 s
Speed = 1 m/s
Distance = Speed × Time = 1 × 700 = 700 m
Number of rounds covered in 200 m = 1
Number of rounds covered in 1 m =
Number of rounds covered in 700 =
Number of rounds covered in 700 = = 3.5 rounds
Let ABCD be a square. Start from point A.
After completing 3 rounds, the player reached A point.
After completing 3.5 rounds, the player reached C point.
We know that, displacement is the shortest distance between the initial and final point.
So, the shortest path here for the player is distance AC.
This leads to the formation of a Δ ABC which is right-angled at B (being angle of a square).
Here, AC = Hypotenuse (being side opposite to right angle)
AB = Base = 50 m
BC = Perpendicular = 50 m
Using Pythagoras Theorem Result i.e.
(Hypotenuse)² = (Perpendicular)² + (Base)², we get,
(AC)² = (BC)² + (AB)²
AC² = 50² + 50²
AC² = 2 × 50²
AC =
AC = 50√2 m
∴ The magnitude of of displacement of the farmer at the end of 11 min 40 sec from his initial position is 50√2 m.
