# A particle starts sliding down a friction less inclined plane. if Sn is distance travelled by it from time t=n-1 sec to n sec then find the ratio Sn/Sn+1. .... plz explain.....

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first, find distance travelled in n seconds,

s=ut+1/2gt^2

here, u=0, t=n.

then find distance travelled in n-1 seconds.

the difference between those two is Sn.

Sn+1 denotes the distance that is travelled by particle from time n seconds to n+1 seconds.

as you have found Sn, find Sn+1.

then you get the ratio.

if you are still facing problems in understanding, ask where you need the details.

happy to help! :)

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Use the equations of motion: s = ut + 1/2 a t^2

let Ф be the angle of the inclined plane with the horizontal.

Acceleration along the inclined plane = g sin Ф

Let initial velocity at t = 0 sec, be u = 0 m/s.

distance travelled up to n-1 sec: D(n-1) = 1/2 g SinФ * (n-1)²

distance travelled up to n sec. : D(n) = 1/2 * g Sin Ф * n²

Distance travelled upto n+1 sec: D(n+1) = 1/2 * g Sin Ф * (n+1)²

Sn = Distance travelled from t = n -1 sec. to n sec. = 1/2 g Sin Ф * [n² - (n-1)²]

= 1/2 g Sin Ф * (2n - 1)

Sn+1 = distance travelled from t = n sec to n+1 sec = 1/2 g Sin Ф * (2n +1)

Ratio : Sn / Sn+1 = (2n -1) / (2n +1) = 1 - 1 / (n +1/2)

let Ф be the angle of the inclined plane with the horizontal.

Acceleration along the inclined plane = g sin Ф

Let initial velocity at t = 0 sec, be u = 0 m/s.

distance travelled up to n-1 sec: D(n-1) = 1/2 g SinФ * (n-1)²

distance travelled up to n sec. : D(n) = 1/2 * g Sin Ф * n²

Distance travelled upto n+1 sec: D(n+1) = 1/2 * g Sin Ф * (n+1)²

Sn = Distance travelled from t = n -1 sec. to n sec. = 1/2 g Sin Ф * [n² - (n-1)²]

= 1/2 g Sin Ф * (2n - 1)

Sn+1 = distance travelled from t = n sec to n+1 sec = 1/2 g Sin Ф * (2n +1)

Ratio : Sn / Sn+1 = (2n -1) / (2n +1) = 1 - 1 / (n +1/2)