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

Correct Answer: 19

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:

\[ \frac{2}{x} = 1 - \frac{17}{19} = \frac{2}{19}\]

\[ x = 19 \]