The distance between the highest height of a cabin on a ferris wheel and the cabin's lowest point is $34 \mathrm{~m}$, as shown. If the ferris wheel rotates with angular velocity 0.07 $\mathrm{rad} / \mathrm{s}$, what distance do riders on the ferris wheel travel in 30.0 seconds?

Step 1 ：The distance between the highest height of a cabin on a ferris wheel and the cabin's lowest point is \(34 \mathrm{~m}\).

Step 2 ：The ferris wheel rotates with angular velocity \(0.07 \mathrm{rad} / \mathrm{s}\).

Step 3 ：We are asked to find the distance riders on the ferris wheel travel in \(30.0\) seconds.

Step 4 ：The distance travelled by the riders on the ferris wheel can be calculated by multiplying the angular velocity by the time and the radius of the ferris wheel.

Step 5 ：The radius of the ferris wheel is half of the distance between the highest and lowest point of the cabin, which is \(17.0 \mathrm{~m}\).

Step 6 ：Substituting the values into the formula, we get \(d = 0.07 \times 30.0 \times 17.0 = 35.7 \mathrm{~m}\).

Step 7 ：Final Answer: The riders on the ferris wheel travel a distance of \(\boxed{35.7 \mathrm{~m}}\) in 30.0 seconds.