I Wish The Car Stayed Still

You can represent the ladybug's position as a vector. You can decompose that vector into two vectors: one vector that connects the origin and the midpoint of diameter on the $x$ -axis on the Volkswagen Beetle, and another that connects the aforementioned midpoint and the ladybug's position.

The vector connecting the origin and aforementioned midpoint is $\left \langle 10t, 0 \right \rangle$ because it is mentioned that its speed is 10. The vector connecting aforementioned midpoint and the ladybug's position is $\left \langle \cos t, \sin t \right \rangle$ because it's a unit semicircle. Adding the vectors together to get the vector giving the ladybug's position gives $\left \langle 10t + \cos t, \sin t \right \rangle$ .

Let's call the vector that gives the ladybug's position $\vec{L}$ . Then , $\dfrac{d\vec{L}}{dt} = \left \langle 10 - \sin t, \cos t \right \rangle$ . But that's a velocity vector. The speed has to be a scalar, which is consequently $\left| \left| \dfrac{d\vec{L}}{dt} \right| \right|$ , or the norm of the velocity vector, which is $\sqrt{(10 - \sin t)^{2} + \cos^{2} t }$ . Let's denote speed as $\dfrac{dS}{dt} = \sqrt{101 - 20 \sin t}$ .

We wanted to know the speed at time $t= \dfrac{\pi}{4}$ , so $\left. \dfrac{dS}{dt} \right|_{t=\frac{\pi}{4}} = \sqrt{101- 20 \sin \dfrac{\pi}{4}}$ .

Finally, the problem statement asks for the floor of 10 times the speed at $t= \dfrac{\pi}{4}$ , which turns out to be $\boxed{93}$ .