Question

A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let S, be the distance travelled by the block in the interval Sn t=n-1 to t=n. Then, the ratio Sn/Sn+1 is:​

Answer 1

Answer:

The answer is : $\frac{S_{n} }{S_{n+1} } =\frac{2n-1}{2n+1}$

Explanation:

$S_{n}$=distance in nth second

$S_{n+1}$=distance in (n+1)th second

t=n to t=n+1

$S_{n} =\frac{a}{2} (2n-1)$  where a=acceleration

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