A small block slides without friction down an inclined plane starting from rest. Let sn be the distance travelled from t = n - 1 to t = n. Then, (Sn/Sn+1) is

Question

A small block slides without friction down an inclined plane starting from rest. Let sn be the distance travelled from t = n - 1 to t = n. Then, (Sn/Sn+1) is (A) (2n-1/2n) (B) (2n+1/2n-1) (C) (2n-1/2n+1) (D) (2n/2n+1)

Tardigrade Admin · Accepted Answer

Distance travelled in t second is, st = u+at - (1/2)a Given, u = 0 ∴ (sn/sn+1)=(an-(1/2)a/a(n+1)-(1)2a)=(2n-1/2n+1) ∴ Correct option is (c).

Q. A small block slides without friction down an inclined plane
starting from rest. Let $s_{n}$ be the distance travelled from
$t = n - 1$ to $t = n$ . Then, $\frac{S _{n}}{S _{n + 1}}$ is

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

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

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

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

Distance travelled in t second is, $s_{t} = u + a t - \frac{1}{2} a$
Given, u = 0
$∴ \frac{s _{n}}{s _{n + 1}} = \frac{an - \frac{1}{2} a}{a ( n + 1 ) - \frac{1}{2} a} = \frac{2 n - 1}{2 n + 1}$
$∴$ Correct option is (c).

Q. A small block slides without friction down an inclined plane starting from rest. Let sn​ be the distance travelled from t=n−1 to t=n. Then, Sn+1​Sn​​ is

2n2n−1​

2n−12n+1​

2n+12n−1​

2n+12n​