A particle moves in a circle of radius 1.0 cm with a speed given by v= 2t, where v is in cm/s and t in seconds.

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the radial acceleration of the particle at t=1s is $4 cm / s^{2}$

the tangential acceleration at t=1s is $2 cm / s^{2}$

the magnitude of net acceleration at t=1s is $2 \sqrt{5} cm / s^{2}$

the magnitude of net acceleration at t=1s $13 \sqrt{5} cm / s^{2}$

answer is A, B, C.

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Detailed Solution

$a_{r} = \frac{v^{2}}{R} = \frac{(2 t)^{2}}{1.0} = 4 t^{2}$

At $t = 1 s, a_{r} = 4 cm / s^{2}$

(b) $a_{t} = \frac{d v}{d t} = 2 cm / s^{2}$

$= 2 \sqrt{5} cm / s^{2}$

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