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Q.

Speed of a particle rotating in a circle of radius 4m is $v = t^{2}$ . Magnitude of its acceleration at t=2sec is

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a

$4 \sqrt{2} m / s^{2}$

b

$8 m / s^{2}$

c

$4 m / s^{2}$

d

$8 \sqrt{2} m / s^{2}$

answer is B.

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Detailed Solution

$v = t^{2}$

$a_{T} = \frac{d v}{d t} = 2 t$

$a_{c} = \frac{v^{2}}{R} = \frac{4 t^{2}}{R}$

$| a | = \sqrt{a_{T}^{2} + a_{c}^{2}}$

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Speed of a particle rotating in a circle of radius 4m is v=t2. Magnitude of its acceleration at t=2sec is