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3-2.Motion in Plane
hard
A particle moves along an arc of a circle of radius $R$ . Its velocity depends on the distance covered as $v = a\sqrt s$ , where $a$ is a constant then the angle $\alpha $ between the vector of the total acceleration and the vector of velocity as a function of $s$ will be
A$\tan \alpha = \frac{R}{{2s}}$
B$\tan \alpha = \frac{2s}{{R}}$
C$\tan \alpha = \frac{2R}{{s}}$
D$\tan \alpha = \frac{s}{{2R}}$
Solution
$\mathrm{V}=\mathrm{a} \sqrt{\mathrm{S}}$
$\frac{d V}{d S}=\frac{9}{2 \sqrt{S}}$
$\mathrm{V} \frac{\mathrm{dV}}{\mathrm{dS}}=\mathrm{a} \sqrt{\mathrm{S}} \cdot \frac{\mathrm{a}}{2 \sqrt{\mathrm{S}}}=\frac{\mathrm{a}^{2}}{2}=$
$=$ tangential component $=\mathrm{q}_{\mathrm{t}}$
$\frac{V^{2}}{R}=a_{C}=\frac{a^{2} S}{R}$
$\therefore \tan \alpha=\frac{a_{c}}{a_{t}}=\frac{2 S}{R}$
$\frac{d V}{d S}=\frac{9}{2 \sqrt{S}}$
$\mathrm{V} \frac{\mathrm{dV}}{\mathrm{dS}}=\mathrm{a} \sqrt{\mathrm{S}} \cdot \frac{\mathrm{a}}{2 \sqrt{\mathrm{S}}}=\frac{\mathrm{a}^{2}}{2}=$
$=$ tangential component $=\mathrm{q}_{\mathrm{t}}$
$\frac{V^{2}}{R}=a_{C}=\frac{a^{2} S}{R}$
$\therefore \tan \alpha=\frac{a_{c}}{a_{t}}=\frac{2 S}{R}$
Standard 11
Physics
