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A particle moves in a circular path of radius $R$ with an angular velocity $ω = a - bt$ , where $a$ and $b$ are positive constants and $t$ is time. The magnitude of the acceleration of the particle after time $\frac{2 a}{b}$ is

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$\frac{a}{R}$

$a^{2} R$

$R (a^{2} + b)$

$R \sqrt{a^{4} + b^{2}}$

answer is D.

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Detailed Solution

$α = \frac{d (ω)}{dt} = - b$

$a_{i} = Rα = - Rb$

$t = \frac{2 a}{b}, ω = - a$ $a_{n} = {Rω}^{2} = {Ra}^{2}$

Now,

$a = \sqrt{a_{t}^{2} + a_{n}^{2}} = R \sqrt{a^{4} + b^{2}}$

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