A particle is moving along a circle such that it completes one revolution in $40$ seconds. In $2$ minutes $20$ seconds, the ratio $\frac{| d i s p l a c e m e n t |}{d i s t a n c e}$ is?

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Solution

2 minutes 20 seconds $= 2 \times 60 + 20 = 120 + 20 = 140 s = 40 \times 3 + 20 s$
After 2 minutes 20 seconds the particle has completed 3 revolutions and a half revolution,therefore is at the other end of the diameter.
Displacement=2r
Distance covered = $3 \times 2 π r + π r = 7 π r$
$\frac{| d i s p l a c e m e n t |}{d i s t a n c e}$ = $\frac{2 r}{7 π r} = \frac{2}{7 π}$

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A particle is moving along a circle such that it completes one revolution in 40 seconds. In 2 minutes 20 seconds, the ratio |displacement|distance is?

2 minutes 20 seconds =2×60+20=120+20=140s=40×3+20sAfter 2 minutes 20 seconds the particle has completed 3 revolutions and a half revolution,therefore is at the other end of the diameter.Displacement=2rDistance covered =3×2πr+πr=7πr|displacement|distance=2r7πr=27π

A particle is moving along a circle such that it completes one revolution in $40$ seconds. In $2$ minutes $20$ seconds, the ratio $\frac{| d i s p l a c e m e n t |}{d i s t a n c e}$ is?