A particle experiences constant acceleration for 20 s after starting from rest. If it travels a distance \[{S_1}\] in the first 10 seconds and distance \[{S_2}\] in the next 10 seconds, then
A. \[{S_2} = {S_1}\]
B. \[{S_2} = 2{S_1}\]
C. \[{S_2} = 3{S_1}\]
D. \[{S_2} = 4{S_1}\]

Hint:Assuming the constant acceleration of the particle as a, calculate the express the distance travelled by the particle in the first 10 seconds using the kinematic equation. Determine the final velocity of the particle at the end of the first 10 seconds. This is the initial velocity for the next interval. Express the distance travelled by the particle in the next 10 seconds and identify the relation between \[{S_1}\] and \[{S_2}\].

Formula used:
\[S = ut + \dfrac{1}{2}a{t^2}\]
Here, S is the displacement, u is the initial velocity, a is the final velocity and t is the time.

Complete step by step answer:
We have given that the distance travelled by the particle in the first 10 seconds is \[{S_1}\] and the distance travelled by the particle in the next 10 seconds is \[{S_2}\]. We have constant acceleration of the particle in these two intervals and let the acceleration is $a$. Let’s express the distance travelled by the particle in first 10 seconds using the kinematic equation as follows,
\[{S_1} = u{t_1} + \dfrac{1}{2}at_1^2\]

Since the particle has started from the rest, the initial velocity of the particle is zero. Therefore, we can write the above equation as,
\[{S_1} = \dfrac{1}{2}at_1^2\]
Substituting 10 seconds for \[{t_1}\], we get,
\[{S_1} = \dfrac{1}{2}a{\left( {10} \right)^2}\]
\[ \Rightarrow {S_1} = 50a\] …… (1)

Now, let’s determine the velocity of the particle at the end of first 10 seconds using the kinematic equation as follows,
\[v = u + a{t_1}\]
\[ \Rightarrow v = a{t_1}\]
Substituting 10 sec for \[{t_1}\], we get,
\[v = 10a\]

This final velocity at the end of the first 10 seconds is the initial velocity for the next 10 seconds. Therefore, express the distance travelled by the particle in next 10 sec as follows,
\[{S_2} = u{t_2} + \dfrac{1}{2}at_2^2\]
Substituting \[10a\] for u and 10 sec for \[{t_2}\], we get,
\[{S_2} = \left( {10a} \right)\left( {10} \right) + \dfrac{1}{2}a{\left( {10} \right)^2}\]
\[ \Rightarrow {S_2} = 100a + 50a\]
\[ \Rightarrow {S_2} = 150a\] …… (2)
Dividing equation (2) by equation (1), we get,
\[\dfrac{{{S_2}}}{{{S_1}}} = \dfrac{{150a}}{{50a}}\]
\[ \Rightarrow \dfrac{{{S_2}}}{{{S_1}}} = 3\]
\[ \therefore {S_2} = 3{S_1}\]

So, the correct answer is option C.

Note: In the alternative way, students can solve this question by calculating the total distance travelled by the particle starting from the rest in 20 seconds. Then calculate the distance travelled in the first 10 seconds and subtract it from the total distance travelled. Note that, in our solution, the initial velocity of the particle for the second interval is not zero.