91Ó°ÊÓ

Position vector $\stackrel{→}{r}$is directed at an angle $\mathrm{θ}$from center.

This problem is based on concept on uniform circular motion. It is the motion of a body moving at a constant speed on a circular path.

Formulae:

Position vector can be written as:

$\stackrel{→}{\mathrm{r}}={\mathrm{r}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{r}}_{\mathrm{y}}\hat{\mathrm{j}}$

Where,

$r_x=r\cos \left(\mathrm{θ}\right)$

$r_y=r\sin \left(\mathrm{θ}\right)$

Vertical component of position vector as:

$r_y=r\sin \left(\mathrm{θ}\right)$

And it would be greatest when $\mathrm{θ}=90°$or $270°$, because sine function has max value at$90°$and$70°$

Vertical component of velocity:

$$\begin{gathered} v_y=\frac{d\left(r_y\right)}{dt} \\ =r\cos \left(\mathrm{θ}\right) \end{gathered}$$

And it would be greatest when $\mathrm{θ}=0°\mathrm{or}180°$because cos function has max value at$\mathrm{θ}=0°\mathrm{and}180°$ .

Vertical component of particle’s acceleration:

$$\begin{gathered} a_y=\frac{d\left(v_y\right)}{dt} \\ =-r\sin \left(\mathrm{θ}\right) \end{gathered}$$

Sine function have max value at $90°\mathrm{and}270°$. Hence, acceleration would be max at$\mathrm{θ}=90°\mathrm{or}270°$ .