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A particle rotates in a circular path of radius $54 m$ with varying speed $v = 4 t^{2} .$ Here $v$ is in $m / s$ and $v$ in second. Find angle between velocity and acceleration $a t = 3 s$ .

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$45 °$

$65 °$

$95 °$

$50 °$

answer is A.

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Detailed Solution

$a_{t} = \frac{d v}{d t} = 8 t$

$a_{r} = \frac{v^{2}}{R} = \frac{{(4 t^{2})}^{2}}{54} = \frac{8}{27} t^{4}$

$At t = 3 s$

$a_{t} = 24 m / s^{2}$

$a_{r} = 24 m / s^{2}$

$\tan θ = \frac{a_{t}}{a_{r}} = 1$

$∴ θ = 45 °$

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