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A motor cycle starts from rest and accelerates along a straight path at $2 \mathrm{~m} / \mathrm{s}^2$. At the starting point of the motor cycle there is a stationary electric sire. How far has the motor cycle gone when the driver hears the frequency of the siren at $94 \%$ of its value when the motor cycle was at rest? (speed of sound $\left.=330 \mathrm{~ms}^{-1}\right)$
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The correct answer is:
$98 \mathrm{~m}$
$98 \mathrm{~m}$
Motor cycle, $u=0, a=2 \mathrm{~m} / \mathrm{s}^2$
Observer is in motion and source is at rest.
$$
\begin{aligned}
& \Rightarrow n^{\prime}=n \frac{v-v_O}{v+v_S} \Rightarrow \frac{94}{100} n=n \frac{330-v_O}{330} \Rightarrow 330-v_0=\frac{330 \times 94}{100} \\
& \Rightarrow v_O=330-\frac{94 \times 33}{10}=\frac{33 \times 6}{10} \mathrm{~m} / \mathrm{s} \\
& s=\frac{v^2-u^2}{2 a}=\frac{9 \times 33 \times 33}{100}=\frac{9 \times 1089}{100} \approx 98 \mathrm{~m} .
\end{aligned}
$$
Observer is in motion and source is at rest.
$$
\begin{aligned}
& \Rightarrow n^{\prime}=n \frac{v-v_O}{v+v_S} \Rightarrow \frac{94}{100} n=n \frac{330-v_O}{330} \Rightarrow 330-v_0=\frac{330 \times 94}{100} \\
& \Rightarrow v_O=330-\frac{94 \times 33}{10}=\frac{33 \times 6}{10} \mathrm{~m} / \mathrm{s} \\
& s=\frac{v^2-u^2}{2 a}=\frac{9 \times 33 \times 33}{100}=\frac{9 \times 1089}{100} \approx 98 \mathrm{~m} .
\end{aligned}
$$
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