A toy cyclist completes one round of a square track of side 2 m in 40 seconds. What will be the displacement at the end of 3 minutes?
- 52 m
- zero
- 16 m
- $2 \sqrt{2}$ m
The Correct Option is D
Solution and Explanation
Total time of motion on square track
= 3 min = 3 \(\times\) 60 = 180 s
Time period of revolution = 40 s
\(\therefore\) Displacement in 160 s (= 4 \(\times\) 40) = 0 as, toy cyclist will be reaching at the starting point.
Thus displacement in 180 s = displacement in 20 s
= \(\sqrt{(2)^2 + (2)^2 } = 2 \sqrt{2} \, m\)
So, the correct option is (D) : \(2\sqrt2\)
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration
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