Question

Tangential acceleration of a particle moving in a circle of radius $1m$ varies with time $t$ as shown in figure (initial velocity of the particle is zero). Time after which total acceleration of particle makes an angle of $30^{\circ }$ with radial acceleration is

• A. $4s$
• B. $\frac{4}{3}s$
• C. $2^{2/3}s$
• D. $\sqrt{2}$ $s$

Answer 1

Answer: (C)

Given that $\frac{{\text{a}}_{\text{t}}}{\text{t}}={\text{tan 60}}^{\text{o}}$
$\Rightarrow a_t=\sqrt{3}\mathrm{t}$
$\frac{\text{dv}}{\text{dt}}=\sqrt{3}\text{t}$
$\Rightarrow {\int }_0^{\text{v}}\text{dv}=\sqrt{3}\text{t dt}$

$\text{v}=\frac{\sqrt{3}{\text{t}}^2}{2}$ ...(2)
Given that ${\text{tan 30}}^{\text{o}}=\frac{{\text{a}}_{\text{t}}}{{\text{a}}_{\text{c}}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{{\text{a}}_{\text{t}}}{{\text{a}}_{\text{c}}}$
$\Rightarrow {\text{a}}_{\text{c}}=\sqrt{3}{\text{a}}_{\text{t}}$
$\frac{{\mathrm{v}}^2}{\mathrm{R}}=\sqrt{3}{\mathrm{a}}_{\mathrm{t}}$
From (1) and (2) $\frac{{\left(\frac{\sqrt{3}}{2}{t}^2\right)}^2}{1}=\sqrt{3}\times \sqrt{3}\mathrm{t}\Rightarrow \frac{3}{4}{t}^4=3\mathrm{t}$
${\text{t}}^3=4$
$\Rightarrow \text{t}=2^{2/3}\text{sec.}$