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A body accelerates from rest with a uniform acceleration $a$ for a time $t$. The uncertainty in $a$ is $8 \%$ and the uncertainty in $t$ is $4 \%$. The uncertainty in the speed is
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The correct answer is:
$12 \%$
Given, uncertainity in acceleration, $\frac{\Delta a}{a} \times 100=8 \%$
Uncertainity in time, $\frac{\Delta t}{t} \times 100=4 \%$
Since, body starts motion from rest, hence by the first equation of motion,
$$
\begin{aligned}
v &=u+a t \\
v &=0+a t \\
\Rightarrow \quad \frac{\Delta v}{v} &=\frac{\Delta a}{a}+\frac{\Delta t}{t} \\
\Rightarrow \quad \frac{\Delta v}{v} \times 100 &=\frac{\Delta a}{a} \times 100+\frac{\Delta t}{t} \times 100 \\
&=8 \%+4 \%=12 \%
\end{aligned}
$$
Uncertainity in time, $\frac{\Delta t}{t} \times 100=4 \%$
Since, body starts motion from rest, hence by the first equation of motion,
$$
\begin{aligned}
v &=u+a t \\
v &=0+a t \\
\Rightarrow \quad \frac{\Delta v}{v} &=\frac{\Delta a}{a}+\frac{\Delta t}{t} \\
\Rightarrow \quad \frac{\Delta v}{v} \times 100 &=\frac{\Delta a}{a} \times 100+\frac{\Delta t}{t} \times 100 \\
&=8 \%+4 \%=12 \%
\end{aligned}
$$
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