Question

A body moves on a frictionless plane starting from rest. If $S_n$ is the distance moved between $t = n - 1$ and $t = n$ and $S_{n-1}$ is the distance moved between $t = n - 2$ and $t = n - 1$, then the ratio $\frac{S_{n-1}}{S_n}$ is $\left(1 - \frac{2}{x}\right)$ for $n = 10$. The value of $x$ is _________.

Answer 1

The displacement in the n th interval, S n, is given by:

S n = $\frac{1}{2}a(2n - 1) = \frac{19a}{2}$ for n = 10.

Similarly, the displacement in the (n − 1)th interval, S n-1, is:

S n-1 = $\frac{1}{2}a(2n - 3) = \frac{17a}{2}$

The ratio of S n-1 to S n becomes:

$\frac{S_{n-1}}{S_{n}} = \frac{\frac{17a}{2}}{\frac{19a}{2}} = \frac{17}{19}$

Now equating this ratio to $1 - \frac{2}{x}$:

$\frac{17}{19} = 1 - \frac{2}{x}$

Simplify to find x :

$\frac{2}{x} = 1 - \frac{17}{19} = \frac{2}{19}$

$x = 19$