Question

A point moves along a circle with speed $v=at$. The total acceleration of the point at a time when it has traced ${\frac{1}{8}}^{th}$ of the circumference is

• A. $8a$
• B. $2a\sqrt{4+{\pi }^2}$
• C. $a$
• D. $\frac{a}{2}\sqrt{4+{\pi }^2}$

Answer 1

The correct option is D $\frac{a}{2}\sqrt{4+{\pi }^2}$

V = at

$1/8^{th}$ of circumference $\Rightarrow \frac{2\pi R}{8}=\frac{\pi R}{4}$

$S=\frac{1}{2}a{t}^2$

$(\because u=0$ at $t=0)$

$\frac{\pi R}{4}=\frac{1}{2}a{t}^2$

$\Rightarrow t_0=\sqrt{\frac{\pi R}{2a}}$

At time $t=t_0,v_0=at\Rightarrow v_0=\sqrt{\frac{\pi R_0}{2}}$

$a_c=\frac{v_0^2}{R}=\frac{\pi Ra}{2R}$ $a_t=0$

$\Rightarrow a_c=\frac{\pi }{2}a$ $a_t=a$

$a_{nf}\sqrt{a_i^2+a_t^2}=a\sqrt{1+\frac{{\pi }^2}{4}}\Rightarrow a_{net}\frac{a}{2}\sqrt{4+{\pi }^2}$