Circular Motion

We know, $a = r\,{w^2} = r \times {\left( {{{2\pi } \over T}} \right)^2}$

Given, ${T_1} = {T_2} = T$

${a_1} = {r_1} \times {\left( {{{2\pi } \over T}} \right)^2}$

${a_2} = {r_2} \times {\left( {{{2\pi } \over T}} \right)^2}$

$\therefore$ ${{{a_1}} \over {{a_2}}} = {{{r_1}} \over {{r_2}}}$

For a particle in uniform circular motion,

${a_c} = {{{v^2}} \over R}$ towards the center of the circle

From figure, $\overrightarrow a = {a_c}\cos \theta \left( { - \widehat i} \right) + {a_c}\sin \theta \left( { - \widehat j} \right)$

$ = {{ - {v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j$

Given $s = {t^3} + 5 $

$\Rightarrow$ Speed,$\,\,\,v = {{ds} \over {dt}} = 3{t^2}$

Tangential acceleration ${a_t} = {{dv} \over {dt}} = 6t$

Radial acceleration ${a_c} = {{{v^2}} \over R} = {{9{t^4}} \over R}$

At $\,\,\,\,t = 2s,\,\,{a_t} = 6 \times 2 = 12\,\,m/{s^2}$

${a_c} = {{9 \times 16} \over {20}} = 7.2\,\,m/{s^2}$

$\therefore$ Net acceleration

$ = \sqrt {a_t^2 + a_c^2} $

$= \sqrt {{{\left( {12} \right)}^2} + {{\left( {7.2} \right)}^2}} $

$ = \sqrt {144 + 51.84} $

$ = \sqrt {195.84} $

$ = 14\,m/{s^2}$

Only option $(b)$ is false since acceleration vector acts along the radius of the circle or towards center of the circle for uniform circular motion and velocity vector always acts along the tangent of the circle.

For no skidding along curved track,

The maximum velocity possible ${v_{\max }} = \sqrt {\mu rg} $

Here $\mu = 0.6,\,r = 150m,\,g = 9.8$

$\therefore$ ${v_{\max }} = \sqrt {0.6 \times 150 \times 9.8} \simeq 30m/s$