Test Index

Motion in a Plane

Section: Physics

Question : 66 of 94

Marks: +1, -0

v = (0.5 t^{2} \hat{i} + 3 t \hat{j} + 9 \hat{k})\ \text{m/s}

$\tan^{-1}\left(\;\frac{2}{3}\right)$ from $X$ -axis
$\tan^{-1}\left(\;\frac{2}{3}\right)$ from $\gamma$ -axis
$\tan^{-1}\left(\;\frac{5}{2}\right)$ from $Y$ -axis
$\tan^{-1}\left(\;\frac{5}{2}\right)$ from $X$ -axis
None of the above

Solution:

Given, velocity of mosquito,

v = (0.5 t^{2} \hat{i} + 3 t \hat{j} + 9 \hat{k})\ \text{m/s}

t = 2\ \text{s}, v = 0.5(2)^{2} \hat{i} + 3(2) \hat{j} + 9 \hat{k}

= 2 \hat{i} + 6 \hat{j} + 9 \hat{k}

\therefore

Angle along

X

-axis,

\cos \alpha =\; \frac{2}{\sqrt{2^{2}+6^{2}+9^{2}}} =\; \frac{2}{\sqrt{121}} =\; \frac{2}{11}

\alpha = \cos^{-1}\left(\;\frac{2}{11}\right)

Similarly, angle along

Y

-axis,

\cos \beta =\; \frac{6}{11}

\Rightarrow \; \; \beta = \cos^{-1}\left(\;\frac{6}{11}\right)

Angle along

Z

-axis,

\cos \gamma =\; \frac{9}{11}

\Rightarrow \; \; \gamma = \cos^{-1}\left(\;\frac{9}{11}\right)

\therefore

Angle made by the direction of motion of mosquito will be

\cos^{-1}\left(\;\frac{2}{11}\right)

from

X

-axis

= \tan^{-1} \; \frac{\sqrt{117}}{2}

\cos^{-1}\left(\;\frac{6}{11}\right)

from

Y

-axis

= \tan^{-1} \; \frac{\sqrt{85}}{2}

\cos^{-1}\left(\;\frac{9}{11}\right)

from

Z

-axis

= \tan^{-1}\left(\;\frac{\sqrt{40}}{9}\right)

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