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If the velocity of a particle moving along a straight line with uniform acceleration, is given by $\mathrm{V}=(\sqrt{196-16 \mathrm{X}}) \mathrm{ms}^{-1}$, then its acceleration is ( $\mathrm{x}$ is displacement of the particle)
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The correct answer is:
$-8 \mathrm{~ms}^{-2}$
Velocity of a particle is given by
$\begin{aligned}
& v=\sqrt{196-16 x} \Rightarrow v^2=196-16 x \\
& 2 v \frac{d v}{d t}=-16 \frac{d x}{d t} \\
& v \frac{d v}{d t}=-8 v
\end{aligned}$
Acceleration, $a=-8 \mathrm{~m} / \mathrm{s}^2$
$\begin{aligned}
& v=\sqrt{196-16 x} \Rightarrow v^2=196-16 x \\
& 2 v \frac{d v}{d t}=-16 \frac{d x}{d t} \\
& v \frac{d v}{d t}=-8 v
\end{aligned}$
Acceleration, $a=-8 \mathrm{~m} / \mathrm{s}^2$
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