$textbf{Solution: }$

As the problem mentions, the angular position of the rotating wheel is $Delta theta = 32^{circ}$ which is counterclockwise direction. As we know the counterclockwise direction is defined as positive and clockwise is negative for rotational variables. So, Counterclockwise is the positive rotation direction and clockwise is the negative direction. So, the given angular position is positive.

$textbf{Solution: }$

As the problem mentions, the wheels of the car moves from right to left which is counterclockwise direction. As we know the counterclockwise direction is defined as positive and clockwise is negative for rotational variables. So, Counterclockwise is the positive rotation direction and clockwise is the negative direction. So, the given angular position is positive.

As we know the average angular velocity $omega_{avg}$ is given by the angular displacement divided by the time interval which the displacement occurs.

$$
begin{align*}
omega_{avg} &= dfrac{ Delta theta }{ Delta t }
end{align*}
$$

So, the direction of the angular speed is the same as the direction of the angular position. So, the angular velocity is positive as the angular position is positive.

$textbf{Given: }$

The initial angular velocity of the object is $omega_{i} = 1 mathrm{~rad/s}$. The angular acceleration of the object is $alpha_{avg} = – 0.5 mathrm{~rad/s^{2}}$.

$textbf{Required: }$

Finding if the final angular speed increasing or decreasing.

$textbf{Calculation: }$

Solve for the time interval of $1 mathrm{~s}$:

As the textbook mentions, the average angular acceleration is given by the change in angular velocity within a given time interval

$$
begin{align*}
alpha_{avg} &= dfrac{ Delta omega }{ Delta t } \
&= dfrac{ omega_{f} – omega_{i} }{Delta t } \
&= dfrac{ 7.9 mathrm{~rad/s} – 5.2 mathrm{~rad/s} }{ 1.3 mathrm{~s} } \
&= 2.077 mathrm{~rad/s^{2}}
end{align*}
$$

Rearrange and solve for the final angular speed:

$$
begin{align*}
omega_{f} &= alpha_{avg} ~ Delta t + omega_{i} \
&= – 0.5 mathrm{~rad/s^{2}} times 1 mathrm{~s} + 1 mathrm{~rad/s} \
&= 0.5 mathrm{~rad/s}
end{align*}
$$

So, the final speed decreasing for all time intervals since the time interval is always positive.

$textbf{Given: }$

The angular position is $theta = 2 mathrm{~rad}$.

$textbf{Required: }$

Fidning the arc length.

$textbf{Calculation: }$

As we know the relation between the angles and the arc length of any circle is given by the following relation:

$$
begin{align*}
s &= r ~ theta \
&= 2 r
end{align*}
$$

So, the arc length, in this case, is equal to the diameter of the circle.

Solution:
In order to evaluate the angular speed, we use the following relation:
$$begin{align*}
omega &= dfrac{ 2 pi }{ T } \
end{align*}$$
So, the angular speed is only depending on the periodic time.

We know the periodic time of the hour hand is $12 mathrm{~hrs}$, ie. the hand hour completes $360 degree$ in $12 text{ hrs}$, so the angular speed of the hour hand is:
$$omega_h = frac{2 pi}{12 text{ hrs}}$$

On the other hand, the periodic time of the minute hand is $1 mathrm{~hr}$, ie. the minute hour completes $360 degree$ in $1 text{ hr} = 60 text{ min}$, so the angular speed of the minute hand is:
$$omega_m = frac{2 pi}{1 text{ hrs}}$$

Comparing these two values, we can conclude that the angular speed of the hour hand is **less** than the angular speed of the minute hand, and their ratio is:
$$
begin{aligned}
frac{omega_h}{omega_m} &= frac{frac{2pi}{12}}{frac{2 pi}{1}}\
&= frac{1}{12}\

end{aligned}$$
Hence, it follows:
$$boxed{omega_h = frac{1}{12} omega_m}$$