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?

Question

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? (A) 52 m (B) zero (C) 16 m (D) 2 √2 m

Tardigrade Admin · Accepted Answer

Total time of motion on square track
= 3 min = 3

\times

60 = 180 s
Time period of revolution = 40 s

∴

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
=

(2)^{2} + (2)^{2} = 22 m

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?

52 m

zero

16 m

$22$ m

Total time of motion on square track
= 3 min = 3 $\times$ 60 = 180 s
Time period of revolution = 40 s
$∴$ 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
= $(2)^{2} + (2)^{2} = 22 m$