Physics - Circular Motion Question with Solution | TestHub

The linear speed of the particle at$\mathrm{t}=2 \mathrm{s}$is

$\mathrm{v}{\mathrm{=2}}^2=4$$\mathrm{m} {\mathrm{s}}^{-1}$

$\therefore$Radial acceleration${\mathrm{a}}_{\mathrm{r}}=\frac{{\mathrm{v}}^2}{\mathrm{r}}=\frac{{\left(4\right)}^2}{2}=8 \mathrm{m} {\mathrm{s}}^{-2}$

The tangential acceleration is${\mathrm{a}}_{\mathrm{t}}=\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{t}}=2\mathrm{t}$

$\therefore$Tangential acceleration at$\mathrm{t}=2 \mathrm{s}$is

${\mathrm{a}}_{\mathrm{t}}=\left(2\right)\left(2\right)=4 \mathrm{m} {\mathrm{s}}^{-2}$

$\therefore$Net acceleration of at$\mathrm{t}=2 \mathrm{s}$is

$\mathrm{a}=\sqrt{{\left({\mathrm{a}}_{\mathrm{r}}\right)}^2+{\left({\mathrm{a}}_{\mathrm{t}}\right)}^2}=\sqrt{{\left(8\right)}^2+{\left(4\right)}^2}$

Or$\mathrm{a}=\sqrt{80} \mathrm{m} {\mathrm{s}}^{-2}$