A particle moves along a circular path over a horizontal xycoordinate system, at constant speed. At time ${\mathit{t}}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathit{s}$ , it is at point (5.00 m, 6.00 m) with velocity $\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}\mathbf{)}\hat{\mathbf{i}}$ and acceleration in the positive xdirection. At time ${\mathit{t}}_{\mathbf{2}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$ , it has velocity $\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}\mathbf{)}\hat{\mathbf{i}}$and acceleration in the positive direction. What are the (a) x and (b) y coordinates of the center of the circular path if${\mathit{t}}_{\mathbf{2}}\mathbf{-}{\mathit{t}}_{\mathbf{1}}$is less than one period?

1. The coordinates of a particle at $t_1\mathrm{is}P(5.00m,6.00m)$
2. The velocity of the particle at$t_1\mathrm{is}{v}_1=(3.00m/s)\hat{j}$

1. The time instance, $t_2=10.0s$
2. The velocity of the particle at $t_{2}is{v}_2=(-3.00m/s)\hat{j}$

Using the equation for acceleration, we can find the radius of the circular path. The particle is moving with uniform circular motion. Hence, the magnitude of velocity and acceleration is constant. We sketch the figure as per the given condition. Theparticle covers three-quarters of the circular path. We can find period T and then use the relation between velocity, acceleration, and radius and find the radius of the circular path. From the figure, the coordinates of the center is C (x + r, y).

Formula:

$$\begin{gathered} v=r\omega \\ \omega =\frac{2\tau \tau }{T} \end{gathered}$$

According to the figure, the particle covers three quarters of the circumference of circle from $t_1\mathrm{to}{t}_2$.

Hence, $t_2-t_1$is equal to the three quarters of period T.

$$\begin{gathered} t_2-t_1=\frac{3T}{4} \\ T=\frac{4(t_2-t_1)}{3} \\ =\frac{4(10.0s-4.00s)}{3} \\ =8.00s \\ {t}_2-t_1<T \end{gathered}$$

The particle is moving in a circular path with uniform circular motion; hence themagnitude of velocity and acceleration remains the same. The magnitude of velocity is $v=3.00m/s$.

The relation between velocity, acceleration and radius is,

$$\begin{gathered} v=r\omega \\ v=r\frac{2\tau \tau }{T} \\ r=\frac{vT}{2\tau \tau } \\ =\frac{3.00m/s\times 8.00s}{2\tau \tau } \\ =3.82m \end{gathered}$$

When the particle comes from$t_1\mathrm{to}{t}_2$tothe y coordinates of the particle at the center remains same but x is (x+r, y)

Therefore, the x and y coordinates of the particle at the center is C ( 8.82m ,6.00m) .