Question

Aparticle experiences constant acceleration for 20 seconds 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
$\text{A}\text{. }{{s}_{2}}={{s}_{1}}$
$\text{B}\text{. }{{s}_{2}}=2{{s}_{1}}$
$\text{C}\text{. }{{s}_{2}}=3{{s}_{1}}$
$\text{D}\text{. }{{s}_{2}}=4{{s}_{1}}$

Answer 1

Hint: It is given that the particle is in constant acceleration. Therefore, use the kinematic equation. Use the kinematic equation, $s=ut+\dfrac{1}{2}a{{t}^{2}}$ and write the equation for displacement of the particle for first 10 seconds and for whole 20 seconds. Then subtract both to find its displacement for the next 10 seconds. Then you will be able to find the relation between ${{s}_{1}}$ and ${{s}_{2}}$.

Formula used:
$s=ut+\dfrac{1}{2}a{{t}^{2}}$

Complete step by step answer:
When a body is moving with a constant acceleration we can use the kinematic equations. It is given that the particle is experiencing a constant acceleration. Let the acceleration of the particle be a. Let us analyse the given conditions and figure out which kinematic equations to be used.
It is said that the particle starts from rest. This means that at time t = 0, the velocity of the particle was zero.
Then it is said that the particle travels a distance ${{s}_{1}}$ in the first 10 seconds of its motion. Here we will use the kinematic equation, $s=ut+\dfrac{1}{2}a{{t}^{2}}$.
In this equation, s is the displacement of the particle in time t, u is the initial velocity (at time t = 0) of the particle and a is the acceleration of the particle.
For the first 10 seconds, $s={{s}_{1}}$, u=0 and t=10s.
Hence, ${{s}_{1}}=(0)(10)+\dfrac{1}{2}a{{(10)}^{2}}$
$\Rightarrow {{s}_{1}}=\dfrac{1}{2}a(100)=50a$ ….. (i).
Let the displacement of the particle after 20 seconds be s. Let us calculate the displacement (s) of the for 20 seconds and then ${{s}_{2}}=s-{{s}_{1}}$.
After 20 seconds,
$s=(0)(20)+\dfrac{1}{2}a{{(20)}^{2}}$
$\Rightarrow s=\dfrac{1}{2}a(400)=200a$
This implies that ${{s}_{2}}=200a-50a=150a$ …… (ii).
Therefore, The distance travelled by the particle in the next 10 seconds is equal to 150a.
Divide equation (i) and equation (ii).
This gives that,
$\dfrac{{{s}_{1}}}{{{s}_{2}}}=\dfrac{50a}{150a}=\dfrac{1}{3}$
$\Rightarrow {{s}_{2}}=3{{s}_{1}}$
Hence, the correct option is C.

Note: Note that when a body has a constant acceleration in a straight line, the distance travelled by the body and the displacement of the body at any given time are the same. This is because the direction of motion is fixed.