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In a circular motion of a particle the tangential acceleration of the particle is $a_{t} = 9 m / s^{2}$ the particle is initially at rest and radius of the circular path is 4m. Time after which acceleration of the particle makes an angle of $45^{0}$ with the radial acceleration is

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$\frac{1}{3} \sec$

$\frac{2}{3} \sec$

$1 \sec$

$\frac{4}{3} \sec$

answer is B.

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Detailed Solution

$a_{T} = 9 \Rightarrow V = 9 t$

1F the angle between $\vec{a} a n d {\vec{a}}_{R}$ is $45^{0}$ ,

$\begin{array}{l} |a_{R}| = |a_{T}| \\ \frac{V^{2}}{R} = a_{T} \Rightarrow \frac{{(9 t)}^{2}}{4} = 9 \end{array}$

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