Question

A body starting from rest is moving under a constant acceleration up to 20 sec. If it moves ${{S}_{1}}$ distance in the first 10 sec, and ${{S}_{2}}$ distance in the next 10 sec then \[{{S}_{2}}\] will be equal to:
a) ${{S}_{1}}$
b) $2{{S}_{1}}$
c) $3{{S}_{1}}$
d) $4{{S}_{1}}$

Answer 1

Hint: The body is moving with constant acceleration. Hence the speed of the body will keep on changing as time progresses. Hence using Newton’s second kinematic equation for the two different sections of the total distance, we will be able to determine the distance covered in the second section of the journey.
Formula used:
$S=ut+\dfrac{1}{2}a{{t}^{2}}$

Complete answer:
In the question it is given that the body moves ${{S}_{1}}$ distance in the first 10 sec of the total journey. Let us say that the body is moving with acceleration ‘a’ and with initial velocity ‘u’. Then the distance (S) covered by the body in time ‘t’ from Newton’s second kinematic equation is given by,
$S=ut+\dfrac{1}{2}a{{t}^{2}}$
The body is initially at rest. Therefore its initial velocity is zero. Hence ${{S}_{1}}$ is equal to,
$\begin{align}
  & {{S}_{1}}=(0)t+\dfrac{1}{2}a{{(10)}^{2}} \\
 & \Rightarrow {{S}_{1}}=\dfrac{1}{2}a100 \\
 & \therefore \Rightarrow {{S}_{1}}=50a....(1) \\
\end{align}$
Let us say the same body has velocity ‘v’ at time t. hence from Newton’s first kinematic equation we get,
$v-u=at$
For the next time interval of 10 sec, the body will have some initial velocity which is the final velocity of when it covered distance ${{S}_{1}}$ . The initial velocity ‘x’ of the body hence is,
$\begin{align}
  & v-u=at \\
 & \Rightarrow x-0=a(10) \\
 & \therefore x=10a \\
 & \\
\end{align}$
Hence from Newton’s second kinematic equation The distance ${{S}_{2}}$ covered is,
$\begin{align}
  & {{S}_{2}}=xt+\dfrac{1}{2}a{{t}^{2}} \\
 & \Rightarrow {{S}_{2}}=10a(10)+\dfrac{1}{2}a{{(10)}^{2}} \\
 & \Rightarrow {{S}_{2}}=100a+50a \\
 & \Rightarrow {{S}_{2}}=150a \\
 & \because a=\dfrac{{{S}_{1}}}{50} \\
 & \Rightarrow {{S}_{2}}=150\times \dfrac{{{S}_{1}}}{50} \\
 & \therefore {{S}_{2}}=3{{S}_{1}} \\
\end{align}$

Therefore the correct answer of the above question is option c.

Note:
It is to be noted the value of ‘a’ can be obtained from equation 1. The body is accelerating with constant acceleration. If the body moves with random acceleration i.e. it changes with time, then Newton’s equations do not hold valid.