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Question

(B) Given velocity vector: $\overrightarrow{v} = 0.5t^2 \hat{i} + 3t \hat{j} + 9 \hat{k}$.At $t = 2 \, s$,the velocity is $\overrightarrow{v} = 0.5(2)

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None of these

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(B) Given velocity vector: $\overrightarrow{v} = 0.5t^2 \hat{i} + 3t \hat{j} + 9 \hat{k}$.At $t = 2 \, s$,the velocity is $\overrightarrow{v} = 0.5(2)^2 \hat{i} + 3(2) \hat{j} + 9 \hat{k} = 2 \hat{i} + 6 \hat{j} + 9 \hat{k} \, m/s$.The magnitude of the velocity is $|\overrightarrow{v}| = \sqrt{2^2 + 6^2 + 9^2} = \sqrt{4 + 36 + 81} = \sqrt{121} = 11 \, m/s$.The direction cosines are $\cos \alpha = \frac{v_x}{|v|} = \frac{2}{11}$,$\cos \beta = \frac{v_y}{|v|} = \frac{6}{11}$,and $\cos \gamma = \frac{v_z}{|v|} = \frac{9}{11}$.The angle with the $x$-axis is $\alpha = \cos^{-1}(\frac{2}{11})$.The angle with the $y$-axis is $\beta = \cos^{-1}(\frac{6}{11})$.The angle with the $z$-axis is $\gamma = \cos^{-1}(\frac{9}{11})$.None of the given options match these values. Therefore,the correct choice is $B$.

$A$ mosquito is moving with a velocity $\overrightarrow{v} = 0.5t^2 \hat{i} + 3t \hat{j} + 9 \hat{k} \, m/s$. What will be the direction of the mosquito after $t = 2 \, s$?

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