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:

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

\[ x = 19 \]

Answer 2

Step 1: Given setup
A body starts from rest on a frictionless plane with uniform acceleration \(a\). The displacement in \(n^{\text{th}}\) second is denoted by \(S_n\). We are given that \[ S_{n-1}/S_n = \left(1 - \frac{2}{x}\right), \] for \(n = 10\), and we must find \(x.\)

Step 2: Use kinematics formula for displacement in the \(n^{\text{th}}\) second
When a body starts from rest with constant acceleration \(a\), \[ S_n = \text{distance in the } n^{\text{th}} \text{ second} = \frac{1}{2} a (2n - 1). \] Similarly, \[ S_{n-1} = \frac{1}{2} a [2(n - 1) - 1] = \frac{1}{2} a (2n - 3). \]

Step 3: Form the ratio
\[ \frac{S_{n-1}}{S_n} = \frac{2n - 3}{2n - 1}. \] For \(n = 10\), \[ \frac{S_{9}}{S_{10}} = \frac{17}{19}. \]

Step 4: Compare with the given form
We are told that \[ \frac{S_{n-1}}{S_n} = 1 - \frac{2}{x}. \] So, \[ 1 - \frac{2}{x} = \frac{17}{19}. \] Simplify: \[ \frac{2}{x} = 1 - \frac{17}{19} = \frac{2}{19} \quad \Longrightarrow \quad x = 19. \]

Final answer

19