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Q.

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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a

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

b

$1 \sec$

c

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

d

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

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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In a circular motion of a particle the tangential acceleration of the particle is at=9m/s2 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 450 with the radial acceleration is