Question

a particle is moving along a circle such that it completes one revolution in 40 sec .in 2 min 20 sec the ratio of displacement/distance is

Answer 1

Let the diameter of the circle = x
Time taken to complete 1 revolution=40 seconds

Total time taken by the particle = 140 seconds.

It means that particle covers 140/40 rounds in 140 seconds

140/40 = 3.5 rounds

Distance covered = 2×22/7×x/2×3.5
= 22/7x × 3.5
= 11x

Displacement = diameter of the circle= x

Ratio of distance and displacement
= 11x : x
= 11 : 1

Hope this helps you..! :)

Answer 2

1 rev=40sec

1/40rev=1 sec

In 2 min 20sec

We know 1 min =60sec

So2 min =120 sec

120+20=140 sec

1/40×140rev=140 sec= 7/2 rev or 3.5 rev

i.e. it completes 3 full rev. And half rev.

We know that circumference for 1 full circle = 2pie r

And for half= 2 pie r/2= pie r

I.e. 3.5rev= 3×2pie r + pie r= 7 pie r

7 pie r = distance

For displacement we know it completes 3.5 rev which is curcular owiously so when it completes 3 rev. It starts from a point and only where full circular rev. End i.e. from which point it starts there only it ends i.e displacement =0 for 3 rev. But half i.e. 0.5 rev. Is still left or half rev. Distance = pie r but displacement is not pie r uts 2r

so,here the total displacement= 2r

And we r asked about ratio= displacement/ distance which is= 2r/ 7 pie r

We all know pie=22/7

Ratio= 2r×7/ 7×22×r= 1/11