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  4. 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?

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Q. 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?

JIPMERJIPMER 2013Motion in a Plane

Solution:

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
= $\sqrt{(2)^2 + (2)^2 } = 2 \sqrt{2} \, m$