In a circular motion of a particle the tangential acceleration of the particle is given by $a_{t} = 2 t m / s^{2}$ . The radius of the circle described is $4 m$ . The particle is initially at rest. Time after which total acceleration of the particle makes $45^{0}$ with radial acceleration is ___ $s e c$

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Detailed Solution

$a_{t} = \frac{d V}{d t} = 2 t \Rightarrow d V = 2 t d t$

$\Rightarrow \int_{0}^{v} dV = \int_{0}^{t} 2 t d t \Rightarrow V = t^{2}; a_{C} = \frac{V^{2}}{r} = \frac{t^{4}}{r} = \frac{t^{4}}{4}$

$Tan θ = \frac{a_{t}}{a_{C}} \Rightarrow Tan 45^{\circ} = \frac{2 t}{t^{4} / 4} \Rightarrow t = 2 \sec$

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