P01L2022 – Kitty Kinematics

Kitty Kinematics

A very nice kitty is in a large, flat open field. As you watch, the kitty starts at point $A$ and runs $15\ m$ at a constant velocity in a direction $35{}^\circ $ west of north. From there, the kitty changes direction and runs at the same constant speed for $35\ m$ in a direction $25{}^\circ $ west of south arriving at point $B$. It takes the kitty $8$ seconds to get from point $A$ to point $B$ along this route.

1. What is the kitty’s average speed from point $A$ to point $B$?

\boxed{v_{avg}=\frac{D}{\mathrm{\Delta }t}=\frac{50}{8}=6.25\ \ m/s}

2. Find the magnitude and direction (the angle south of west in degrees) of the kitty’s displacement from point $A$ to point $B$.

Based on the figure above, we derive that the distance between points $A$ and $B$ is equal to

\boxed{r=\sqrt{{\left(35{\mathrm{sin} \left(25\right)\ }+15{\mathrm{sin} \left(35\right)\ }\right)}^2+{\left(35{\mathrm{cos} \left(25\right)\ }-15{\mathrm{cos} \left(35\right)\ }\right)}^2}\approx 30.41\ m}

The angle $\theta $ is then given by

\boxed{{\mathrm{tan} \left(\theta \right)\ }=\frac{y_B}{x_B}\ \ \ \ \Rightarrow \ \ \ \ \ \theta ={{\mathrm{tan}}^{-1} \left(\frac{35{\mathrm{cos} \left(25\right)\ }-15{\mathrm{cos} \left(35\right)\ }}{35{\mathrm{sin} \left(25\right)\ }+15{\mathrm{sin} \left(35\right)\ }}\right)\ }\approx 39.71{}^\circ}

3. Find the magnitude and direction (the angle south of west in degrees) of the kitty’s average velocity vector from point $A$ to point $B$.

\boxed{v_{avg}=\frac{r}{\mathrm{\Delta }t}=\frac{30.41}{8}=3.8\ m/s}

\boxed{\theta =39.71{}^\circ}