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A particle starts its motion from rest under the action of a constant force. If the distance covered in first $10 \mathrm{~s}$ is $\mathrm{s}_1$ and that covered in the first $20 \mathrm{~s}$ is $\mathrm{s}_2$, then
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Verified Answer
The correct answer is:
$s_2=4 s_1$
Key Idea If the particle is moving in a straight line under the action of a constant force then distance covered $\mathrm{s}=\mathrm{ut}+\frac{1}{2}$ at $^2$.
Since the body start from rest $\mathrm{u}=0$
$\therefore \quad \mathrm{s}=\frac{1}{2} \mathrm{at}^2$
Now, $\quad \mathrm{s}_1=\frac{1}{2} \mathrm{a}(10)^2$
and
$$
\mathrm{s}_2=\frac{1}{2} \mathrm{a}(20)^2
$$
Dividing Eq. (i) and Eq. (ii), we get
$$
\begin{aligned}
\frac{\mathrm{s}_1}{\mathrm{~s}_2} & =\frac{(10)^2}{(20)^2} \\
\Rightarrow \quad s_2 & =4 \mathrm{~s}_1
\end{aligned}
$$
Since the body start from rest $\mathrm{u}=0$
$\therefore \quad \mathrm{s}=\frac{1}{2} \mathrm{at}^2$
Now, $\quad \mathrm{s}_1=\frac{1}{2} \mathrm{a}(10)^2$
and
$$
\mathrm{s}_2=\frac{1}{2} \mathrm{a}(20)^2
$$
Dividing Eq. (i) and Eq. (ii), we get
$$
\begin{aligned}
\frac{\mathrm{s}_1}{\mathrm{~s}_2} & =\frac{(10)^2}{(20)^2} \\
\Rightarrow \quad s_2 & =4 \mathrm{~s}_1
\end{aligned}
$$
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