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Question

The velocity of a particle is given as

$\overrightarrow{v} = - x\widehat{i} + 2y\widehat{j} - z\widehat{k}m/s$. The magnitude of acceleration at point $(1,2,4)$ is $\_\_\_\_$ $m/s^{2}$.

Options

A.$\sqrt{6}$
B.9
C.$\sqrt{33}$
D.0

Solution

Sol. [Key Concept: Differentiate each velocity component in time, evaluate at the point, then take the magnitude.]

$$\overrightarrow{v} = - x\widehat{i} + 2y\widehat{j} - z\widehat{k} $$

$$\overrightarrow{a} = \frac{d\overrightarrow{v}}{dt} = - \frac{dx}{dt}\widehat{i} + 2\frac{dy}{dt}\widehat{j} - \frac{dz}{dt}\widehat{k} $$

$$= - ( - x)\widehat{i} + 2(2y)\widehat{j} - ( - z)\widehat{k} = x\widehat{i} + 4y\widehat{j} + z\widehat{k} $$$\therefore\overrightarrow{a}$ at $(1,2,4) = \widehat{i} + 8\widehat{j} + 4\widehat{k}$

$$\therefore|\overrightarrow{a}| = 9\text{ }m/s^{2} $$

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