$textbf{Given: }$

The radius of the tire is $r = 32 mathrm{~cm}$. The angular speed of the tires is $78 mathrm{~rad/s}$.

$textbf{Required: }$

Finding the linear speed of the top of each tire.

$textbf{Calculation: }$

Solve for the linear speed on the top of each tire:

Since we solve for the linear speed of the top of each tire, we should use the diameter of the tire instead of the radius of the tire.

As the textbook mentions, the tangential speed of a rotating object is given by

$$
begin{align*}
v &= d ~ omega \
&= 2 r ~ omega \
&= 2 times 0.32 mathrm{~m} times 78 mathrm{~rad/s} \
&= 49.92 mathrm{~m/s}
end{align*}
$$

So, the linear speed of the top of each tire is $49.92 mathrm{~m/s}$.

The linear speed of the top of each tire is $49.92 mathrm{~m/s}$.

$textbf{Calculation: }$

Since the cans roll without slipping, we should use the diameter of each can instead of the radius.

As the textbook mentions, the tangential speed of rotating cans at the top of them is given by

$$
begin{align*}
v &= d ~ omega \
&= 2 r ~ omega \
end{align*}
$$

So, the speed that the wooden plank moves with is $2 r ~ omega$. Also, the linear speed at the center of each can is equal to $r ~ omega$. So, the distance that the center of each can move at the same time that the wooden plank takes to pass the $1 mathrm{~m}$ distance is half of that distance ($0.5 mathrm{~m}$).

The distance that the center of each can move at the same time that the wooden plank takes to pass the $1 mathrm{~m}$ distance is half of that distance ($0.5 mathrm{~m}$).