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(B) Given: Angular acceleration $\alpha = 4 \ rad/s^2$. Initial angular velocity $\omega_0 = 0$.$1$. The angular velocity at time $t$ is given by $\om

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$0.5$

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(B) Given: Angular acceleration $\alpha = 4 \ rad/s^2$. Initial angular velocity $\omega_0 = 0$.$1$. The angular velocity at time $t$ is given by $\omega = \omega_0 + \alpha t = \alpha t$.$2$. The centrifugal (radial) acceleration is $a_c = \omega^2 r = (\alpha t)^2 r = \alpha^2 t^2 r$.$3$. The tangential acceleration is $a_t = \alpha r$.$4$. We are given that the magnitudes are equal: $a_c = a_t$.$5$. Substituting the expressions: $\alpha^2 t^2 r = \alpha r$.$6$. Dividing both sides by $\alpha r$ (assuming $\alpha, r 
eq 0$): $\alpha t^2 = 1$.$7$. Solving for $t$: $t^2 = \frac{1}{\alpha} = \frac{1}{4}$.$8$. Therefore,$t = \sqrt{\frac{1}{4}} = 0.5 \ s$.

$A$ particle at rest starts moving with a constant angular acceleration of $4 \ rad/s^2$ in a circular path. At what time will the magnitudes of its tangential acceleration and centrifugal acceleration be equal (in $s$)?

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