Question

A small block slides without friction down an inclined plane starting from rest. Let ${s_n}$ be the distance travelled from time $t = 1$ to $t = n$. Then is $\dfrac{{{s_n}}}{{{s_{n + 1}}}}$
A. $\dfrac{{2n - 1}}{{2n}}$
B. $\dfrac{{2n + 1}}{{2n - 1}}$
C. $\dfrac{{2n - 1}}{{2n + 1}}$
D. $\dfrac{{\left( {n - 1} \right)(n + 1)}}{{{n^2} + 2n}}$

Answer 1

To find the value of $\dfrac{{{s_n}}}{{{s_{n + 1}}}}$, we need to find the values of ${s_n}$ and ${s_{n + 1}}$ . Here you will need to use the second equation of motion, use this equation to find the value of ${s_n}$ and ${s_{n + 1}}$. Then use these values to find the value of $\dfrac{{{s_n}}}{{{s_{n + 1}}}}$.

Complete step by step solution:
Given, total distance travelled by a small block is ${s_n}$ when initial time is $t = 1$ and final time is $t = n$. We are asked to find the value of $\dfrac{{{s_n}}}{{{s_{n + 1}}}}$
${s_{n + 1}}$ will be the distance travelled by the block when the initial time is $t = 1$ and final time is $t = n + 1$.
Here, we will use second equation of motion, which is
$s = ut + \dfrac{1}{2}a{t^2}$ ………………….(i)
where $s$ is the distance covered, $u$ is the initial velocity, $t$ is the time taken and $a$ is the acceleration of a body.
Here, it is said that the block starts from rest, which means the initial velocity of the block is zero, that is, $u = 0$
Putting $u = 0$in equation (i) we get,
$s = \dfrac{1}{2}a{t^2}$ ……………...(ii)
Now, we apply second equation of motion for distance ${s_n}$ and ${s_{n + 1}}$

For, distance ${s_n}$the initial time is $t = 1$ and final time is $t = n$
The distance travelled in time $t = 1$ will be (using equation (ii))
$s = \dfrac{1}{2}a{\left( 1 \right)^2}$ …………….(iii)
The distance travelled in time $t = n$ will be (using equation (ii))
$s' = \dfrac{1}{2}a{\left( n \right)^2}$ ………………...(iv)
We can get the distance ${s_n}$ by subtracting equation (iii) from (iv),
${s_n} = s' - s$
Putting the values of $s'$ and $s$ we get,
${s_n} = \dfrac{1}{2}a{\left( n \right)^2} - \dfrac{1}{2}a{\left( 1 \right)^2}$
$\Rightarrow {s_n} = \dfrac{1}{2}a\left( {{n^2} - 1} \right)$ …………………..(v)

For, distance ${s_{n + 1}}$ the initial time is $t = 1$ and final time is $t = n + 1$
The distance travelled in time $t = 1$ will be (using equation (ii))
$s = \dfrac{1}{2}a{\left( 1 \right)^2}$ …………...(vi)
The distance travelled in time $t = n + 1$ will be (using equation (ii))
$s'' = \dfrac{1}{2}a{\left( {n + 1} \right)^2}$ ……………..(vii)
We can get the distance ${s_{n + 1}}$ by subtracting equation (iii) from (iv),
${s_{n + 1}} = s'' - s$
Putting the values of $s''$ and $s$ we get,
${s_{n + 1}} = \dfrac{1}{2}a{\left( {n + 1} \right)^2} - \dfrac{1}{2}a$
$\Rightarrow {s_{n + 1}} = \dfrac{1}{2}a\left( {{{\left( {n + 1} \right)}^2} - 1} \right)$ ………...(viii)

Now, dividing equation (v) by (viii) we get,
$\dfrac{{{s_n}}}{{{s_{n + 1}}}} = \dfrac{{\dfrac{1}{2}a\left( {{n^2} - 1} \right)}}{{\dfrac{1}{2}a{{\left( {n + 1} \right)}^2} - 1}}$
$\Rightarrow \dfrac{{{s_n}}}{{{s_{n + 1}}}} = \dfrac{{\left( {{n^2} - 1} \right)}}{{{{\left( {n + 1} \right)}^2} - 1}}$
$\Rightarrow \dfrac{{{s_n}}}{{{s_{n + 1}}}} = \dfrac{{\left( {n - 1} \right)(n + 1)}}{{{n^2} + 1 + 2n - 1}}$
$\therefore \dfrac{{{s_n}}}{{{s_{n + 1}}}} = \dfrac{{\left( {n - 1} \right)(n + 1)}}{{{n^2} + 2n}}$

Hence, option (D) is the correct answer.

Note: There are three equations of motion, which are
(i) First equation of motion:- $v = u + at$
(ii) Second equation of motion:- $s = ut + \dfrac{1}{2}a{t^2}$
(iii) Third equation of motion:- ${v^2} - {u^2} = 2as$
In the above equations, $u$ is the initial velocity, $v$ is the final velocity, $a$ is the acceleration, $t$ is the time taken and $s$ is the distance covered by a body.