This question aims to calculate for these values: the speed of the blue dot that has been painted on the tread of the rear tire when it is 0.80 m above the road, the angular speed of the tires, and the speed of the blue dot when it is 0.40 m above the road.

Speed is defined as the change in the position of the object with respect to time. In other words, it can also be regarded as the ratio of distance covered to time. It is a scalar quantity. Mathematically, it can be written as:

\[ Speed = \dfrac{Distance covered}{time} \]

\[ S = \dfrac{v}{t} \]

Angular speed is defined as the change in angular displacement with respect to time. A body undergoing circular motion has angular speed. It can be expressed as:

\[ Angular speed = \dfrac{Angular Displacement}{time} \]

\[ \omega = \dfrac{\Theta} {t} \]

**Expert Answer:**

Given:

Diameter of tire d = 0.80 m

Velocity of the bicycle v = 5.6 m/s

In order to calculate the velocity of the blue dot at $0.80 m$ above the ground, the following equation will be used:

\[ v_b = v + r\omega ( eq 1) \]

Where $\omega$ is the angular velocity.

For calculating $\omega$, use the following equation:

\[ \omega = \dfrac{v}{r} \]

Where $r$ is the radius that is given as:

\[ radius = \dfrac{diameter}{2}\]

\[ r = \dfrac{0.80}{2}\]

\[ r = 0.40 \]

So angular velocity is given as:

\[ \omega = \dfrac{5.6} {0.4} \]

\[ \omega = 14 rad/s \]

**Numerical Results:**

Now, putting in the $eq 1$ gives the velocity of the blue dot.

\[ v_b = 5.6 + (0.4)(14) \]

\[ v_b = 11.2 m/s \]

Therefore, the velocity of the blue dot is $11.2 m/s$, and the angular velocity $\omega$ is $14 rad/s$.

**Alternative Solution:**

The angular velocity of the tire is $14 rad/s$.

The speed of the bike’s blue dot when it is $0.80 m$ above the road is given as the sum of its velocity of the center of mass of the wheel and linear velocity of the bicycle.

\[ v_b = v + r\omega \]

\[ v_b = 5.6 + (0.4)(14) \]

\[ v_b = 11.2 m/s \]

**Example:**

A bicycle with $0.80 m$ diameter tires is coasting on a level road at $5.6 m/s$. A small blue dot has been painted on the tread of the rear tire. What is the speed of the bike’s blue dot when it is $0.40 m$ above the road?

The speed of the bike’s blue dot when it is $0.40 m$ above the road can be determined using Pythagoras theorem.

\[ (v_b)^2 = (v)^2 + (r\omega)^2 \]

\[ v_b = \sqrt{(v)^2 + (r\omega)^2} \]

The angular velocity $\omega$ of the tires is given as:

\[ \omega = \dfrac{v}{r} \]

\[ \omega = \dfrac{5.6}{0.4} \]

\[ \omega = 14 m/s \]

Putting in the above equation gives us the speed of blue dot above $0.40 m$.

\[ v_b = \sqrt{(5.6)^2 + (0.4×14)^2} \]

\[ v_b = 7.9195 m/s \]