A particle moves towards east with a velocity | vedclass

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Question

(B) Initial velocity $\vec{v}_1 = 5\hat{i}\ m/s$.Final velocity $\vec{v}_2 = 5\hat{j}\ m/s$.Change in velocity $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}\ m/s^2$ North-West

Vedclass · Answer

(B) Initial velocity $\vec{v}_1 = 5\hat{i}\ m/s$.Final velocity $\vec{v}_2 = 5\hat{j}\ m/s$.Change in velocity $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1 = 5\hat{j} - 5\hat{i}$.The magnitude of change in velocity is $|\Delta\vec{v}| = \sqrt{(-5)^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\ m/s$.The direction of $\Delta\vec{v}$ is North-West.Average acceleration $\vec{a}_{avg} = \frac{\Delta\vec{v}}{\Delta t} = \frac{5\sqrt{2}}{10} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\ m/s^2$.Thus,the average acceleration is $\frac{1}{\sqrt{2}}\ m/s^2$ in the North-West direction.

$A$ particle moves towards east with a velocity of $5\ m/s$. After $10\ s$,its direction changes towards north with the same velocity. The average acceleration of the particle is:

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