Question

A body starting from rest with uniform acceleration travels distance \[{{s}_{1}}\] in the first $t$ second and travels a distance ${{s}_{2}}$ with uniform velocity in the next $2t$ second. Then:
$\begin{align}
  & \text{A}\text{. }{{s}_{2}}=4{{s}_{1}} \\
 & \text{B}\text{. }{{s}_{2}}=2{{s}_{1}} \\
 & \text{C}\text{. }{{s}_{1}}=4{{s}_{2}} \\
 & \text{D}\text{. }{{s}_{1}}=2{{s}_{2}} \\
\end{align}$

Answer 1

Hint: Take the initial velocity of the body to be zero and apply equation of motion to calculate the acceleration in terms of velocity and put this acceleration value in the second equation of motion to get the first displacement \[{{s}_{1}}\] in terms of final velocity of the first part. This final velocity is the uniform velocity in the second part and because the velocity is uniform so the acceleration in the second part is zero. So apply the second equation of motion to get the displacement in the second part. Divide the displacement obtained in both the parts to get the relation among them.

Formulas used:
First equation of motion, $v=u+at$
Where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration during the motion and $t$ is the total time taken.
Second equation of motion,$s=ut+\dfrac{1}{2}a{{t}^{2}}$
Where $s$ is the displacement, $u$ is the initial velocity, $a$ is the acceleration during the motion and $t$ is the total time taken.

Complete answer:
For the first part the body starts from rest with uniform acceleration and travels distance \[{{s}_{1}}\] in the first $t$ second. Let the initial velocity of the object is $u$, the constant acceleration is $a$ and the final velocity be $v$. So
$v=u+at$
As the object starts from rest $u=0$, so
$v=at\text{ or, }a=\dfrac{v}{t}$
But the displacement during the first part is given by
${{s}_{1}}=ut+\dfrac{1}{2}a{{t}^{2}}=\dfrac{1}{2}a{{t}^{2}}\left( \because u=0 \right)$
But $a=\dfrac{v}{t}$,
${{s}_{1}}=\dfrac{1}{2}\dfrac{v}{t}{{t}^{2}}=\dfrac{1}{2}vt$
The final velocity of the first part will be the uniform velocity (as well as initial velocity) in the second part. Also as the velocity is uniform, the acceleration during the second part will be zero. So the displacement during the second part is given by
${{s}_{2}}=vt'+\dfrac{1}{2}at{{'}^{2}}=vt'=v(2t)=2vt$
Now
$\begin{align}
  & \dfrac{{{s}_{1}}}{{{s}_{2}}}=\dfrac{\dfrac{1}{2}vt}{2vt}=\dfrac{1}{4} \\
 & \Rightarrow {{s}_{2}}=4{{s}_{1}} \\
\end{align}$

So the correct option is A.

Note:
Note that the equation of motion we used here is only valid for uniformly accelerated motion. These equations will not hold good for a non-uniformly accelerated motion. Because for a non-uniformly accelerated motion we cannot associate an inertial frame of reference. And for a non-inertial frame of reference Newton's law is not valid.