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(C) Given: Initial velocity $\vec{v}_1 = 6 \hat{i} \, m/s$. Final velocity $\vec{v}_2 = 6(\cos 60^{\circ} \hat{i} + \sin 60^{\circ} \hat{j}) = 6(\frac

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(C) Given: Initial velocity $\vec{v}_1 = 6 \hat{i} \, m/s$. Final velocity $\vec{v}_2 = 6(\cos 60^{\circ} \hat{i} + \sin 60^{\circ} \hat{j}) = 6(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) = 3 \hat{i} + 3\sqrt{3} \hat{j} \, m/s$. Change in velocity $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1 = (3 \hat{i} + 3\sqrt{3} \hat{j}) - 6 \hat{i} = -3 \hat{i} + 3\sqrt{3} \hat{j} \, m/s$. Magnitude of change in velocity $|\Delta \vec{v}| = \sqrt{(-3)^2 + (3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6 \, m/s$. Time interval $\Delta t = 6 \, s$. Average acceleration $a_{av} = \frac{|\Delta \vec{v}|}{\Delta t} = \frac{6}{6} = 1 \, m/s^2$.

$A$ particle is moving eastwards with a speed of $6 \, m/s$. After $6 \, s$,the particle is found to be moving with the same speed in a direction $60^{\circ}$ north of east. The magnitude of average acceleration in this interval of time is ....... $m/s^2$.

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