A particle initially at rest starts moving from reference point. mathrm{x}=0 along mathrm{x} -axis,

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2.Motion in Straight Line

hard

A particle initially at rest starts moving from reference point. $\mathrm{x}=0$ along $\mathrm{x}$-axis, with velocity $v$ that varies as $v=4 \sqrt{\mathrm{x} }\ m/ \mathrm{s}$. The acceleration of the particle is __________$ \mathrm{ms}^{-2}$.

A$7$

B$8$

C$9$

D$10$

(JEE MAIN-2024)

Solution

$\mathrm{V}=4 \sqrt{\mathrm{x}}$
$\mathrm{a}=\mathrm{V} \frac{\mathrm{dv}}{\mathrm{dx}}$
$=4 \sqrt{\mathrm{x}} \times 4 \times \frac{1}{2} \mathrm{x}^{-1 / 2}=8 \mathrm{~m} / \mathrm{s}^2$

Standard 11

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Solution

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The position of a particle moving along the $X-$axis at certain times is given below :Which of the following describes the motion correctly

$\begin{array}{|c|c|c|c|c|} \hline t( s ) & 0 & 1 & 2 & 3 \\ \hline x ( m ) & -2 & 0 & 6 & 16 \\ \hline \end{array} $