A particle starts sliding down a frictionless inclined plane. If $S_{n}$ is the distance traveled by it from time $t =$ n-1 sec to $t =$ n sec, the ratio $S_{n} / S_{n + 1}$ is :

\frac{2 n - 1}{2 n + 1}

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\frac{2 n + 1}{2 n}

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\frac{2 n}{2 n + 1}

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\frac{2 n + 1}{2 n - 1}

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Solution

The correct option is A $\frac{2 n - 1}{2 n + 1}$
Distance travelled by the particle in nth second is given by

$S_{n t h}$ = $\frac{(2 n - 1) a}{2}$

Where $n$ is the nth second and
$a$ is the acceleration of the particle

$S_{n + 1 t h}$ = $\frac{(2 (n + 1) - 1) a}{2}$

$S_{n t h}$ / $S_{n + 1 t h}$ = $\frac{(2 n - 1)}{(2 (n + 1) - 1)}$

Ratio = $\frac{(2 n - 1)}{(2 n + 1)}$

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A particle starts sliding down a frictionless inclined plane. If S_n is the distance travelled by it from time t=n-1 secto t=n sec, find the ratio of S_n/S_n+1.

Explain in detail.

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A particle starts sliding down a frictionless inclined plane. If Sn is the distance traveled by it from time t= n-1 sec to t= n sec, the ratio Sn/Sn+1 is :

The correct option is A 2n−12n+1Distance travelled by the particle in nth second is given bySnth = (2n−1)a2Where n is the nth second and a is the acceleration of the particleSn+1th = (2(n+1)−1)a2Snth/Sn+1th = (2n−1)(2(n+1)−1)Ratio = (2n−1)(2n+1)

A particle starts sliding down a frictionless inclined plane. If $S_{n}$ is the distance traveled by it from time $t =$ n-1 sec to $t =$ n sec, the ratio $S_{n} / S_{n + 1}$ is :