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

Hint: 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.