91Ó°ÊÓ

The time interval, $t_1=200\mathrm{s}$

The velocity of the cat at $t_1\mathrm{is}{\stackrel{â‡\P Ä}{v}}_1=\left(3.00\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}}+\left(4.00\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}$

The time interval, $t_2=5.00\mathrm{s}$

The velocity of the cat at role="math" localid="1657000200311" $t_2\mathrm{is}{\stackrel{â‡\P Ä}{\mathrm{v}}}_2=\left(-3.00\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}}+\left(-4.00\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}$

$t_2-t_1<T$

Using the equation for centripetal acceleration, find the radius of the circular path. The particle is moving with uniform circular motion. Hence, the magnitude of velocity is constant. Sketch the figure as per the given condition where the particle covers half-quarters of the circular path. Find period T and then use the relation between velocity, acceleration, and radius and find the radius of the circular path and then the magnitude of centripetal acceleration. To find the average acceleration we can use the expression of acceleration in vector form and then find its magnitude.

Formula:

$$\begin{gathered} v=\mathrm{Ó\neg } \\ \mathrm{Ó\neg }=\frac{2\mathrm{Ï„}\mathrm{Ï„}}{T} \\ {a}_c=\frac{v^2}{r} \\ \stackrel{â‡\P Ä}{a}=\frac{\stackrel{â‡\P Ä}{a_2}-\stackrel{â‡\P Ä}{v_1}}{t_2-t_1} \end{gathered}$$

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

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

$$\begin{gathered} t_2-t_1=\frac{T}{2} \\ T=2\left(t_2-t_1\right) \\ =2\left(5.00\mathrm{s}-2.00\mathrm{s}\right) \\ =6.00\mathrm{s} \\ {\mathrm{t}}_2-{\mathrm{t}}_1<T \end{gathered}$$

The particle is moving in circular path with uniform circular motion; hence the magnitude of velocity and acceleration remains the same. The magnitude of velocity is

$$\begin{gathered} v=\sqrt{{\left(v_1\right)}^2+{\left(v_2\right)}^2} \\ =\sqrt{{\left(3.00\mathrm{m}/\mathrm{s}\right)}^2+{\left(4.00\mathrm{m}/\mathrm{s}\right)}^2} \\ =5.00\mathrm{m}/\mathrm{s} \end{gathered}$$

The relation between velocity, acceleration and radius is,

$$\begin{gathered} v=\mathrm{Ó\neg } \\ =r\frac{2\mathrm{Ï„}\mathrm{Ï„}}{T} \\ r=\frac{vT}{2\mathrm{Ï„}\mathrm{Ï„}} \\ =\frac{\left(5.00\mathrm{m}/\mathrm{s}\right)×\left(6.00\mathrm{s}\right)}{2\mathrm{Ï„}\mathrm{Ï„}} \\ =4.774\mathrm{m} \end{gathered}$$

The expression of the centripetal acceleration is,

$$\begin{gathered} a_c=\frac{v^2}{r} \\ =\frac{{\left(5.00\mathrm{m}/\mathrm{s}\right)}^2}{4.774\mathrm{m}} \\ =5.24\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The magnitude of the cat’s centripetal acceleration is $5.24\mathrm{m}/{\mathrm{s}}^2$.

The expression of average acceleration is

$$\begin{gathered} \stackrel{â‡\P Ä}{a}=\frac{{\stackrel{â‡\P Ä}{v}}_2-{\stackrel{â‡\P Ä}{v}}_1}{t_2-t_1} \\ \stackrel{â‡\P Ä}{a}=\frac{\left(-3.00\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}}+\left(-4.00\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}\left(3.00\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}}+\left(4.00\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}}{5.00-2.00} \\ =\frac{\left(-6.00\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{i}}+\left(-8.00\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{j}}}{3.00} \\ =\left(-2.00\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{i}}+\left(\frac{-8.00}{3.00}\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{j}} \end{gathered}$$

The magnitude of the average acceleration is,

$$\begin{gathered} a=\sqrt{{\left(a_1\right)}^2+{\left(a_2\right)}^2} \\ =\sqrt{{\left(-2.00\mathrm{m}/{\mathrm{s}}^2\right)}^2+{\left(-8.00/3.00\mathrm{m}/{\mathrm{s}}^2\right)}^2} \\ =3.33\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the magnitude of acceleration is $3.33\mathrm{m}/{\mathrm{s}}^2$.