A particle moves towards east with velocity 5 | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Initial velocity $\vec{v}_i = 5\hat{i}\,m/s$.Final velocity $\vec{v}_f = 5\hat{j}\,m/s$.Change in velocity $\Delta \vec{v} = \vec{v}_f - \vec{v}_i = 5\hat{j} - 5\hat{i}$.Magnitude of change in velocity $|\Delta \vec{v}| = \sqrt{(-5)^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\,m/s$.The direction of $\Delta \vec{v}$ is North-West (since it points in the direction of $-\hat{i} + \hat{j}$).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 velocity $5\, m/s$. After $10\, s$,its direction changes towards north with the same velocity. The average acceleration of the particle is

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