91Ó°ÊÓ

First, we need to find the distance traveled in each part of the trip. Let's denote the distance in the first part as \(d_1\) and the second part as \(d_2\). We are given the speeds and travel times for both parts of the trip. For the first part: Speed, \(v_1 = 108\, \mathrm{km/h} \) Time, \(t_1 = 20.0 \, \mathrm{min} = \frac{1}{3} \, \mathrm{h}\) (Converting minutes to hours) So, the distance for the first part: \(d_1 = v_1 t_1 = 108 \times \frac{1}{3} = 36 \, \mathrm{km}\) For the second part: Speed, \(v_2 = 90.0\, \mathrm{km/h}\) Time, \(t_2 = 10.0 \, \mathrm{min} = \frac{1}{6} \, \mathrm{h}\) (Converting minutes to hours) So, the distance for the second part: \(d_2 = v_2 t_2 = 90 \times \frac{1}{6} = 15 \, \mathrm{km}\) Now, we can find the total distance traveled: \(d = d_1 + d_2 = 36 + 15 = 51 \, \mathrm{km}\) And the total time taken: \(t = t_1 + t_2 = \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \, \mathrm{h}\) Therefore, the average speed is: \( \mathrm{Average \, Speed} = \frac{d}{t} = \frac{51}{0.5} = 102 \, \mathrm{km/h}\)

To find the average velocity, first, we need to find the displacement for each part of the trip separately and then sum them up. For the first part of the trip, the speedboat heads west, so the displacement vector can be represented as: \(\begin{pmatrix}-36 \\ 0 \end{pmatrix} \, \mathrm{km}\) (Negative value along the x-axis represents going west) For the second part of the trip, the speedboat heads \(60^{\circ}\) south of west. So the displacement vector can be represented as: \(\begin{pmatrix}-d_2 \cos(60^\circ) \\ -d_2 \sin(60^\circ) \end{pmatrix} \, \mathrm{km}\) (Negative values represent going west and south) Substituting \(d_2 = 15\, \mathrm{km}\): \(\begin{pmatrix}-15 \times \frac{1}{2} \\ -15 \times \frac{\sqrt{3}}{2} \end{pmatrix} = \begin{pmatrix}-7.5 \\ -12.99 \end{pmatrix} \, \mathrm{km}\) Summing the displacements for both parts: \(\begin{pmatrix}-36 \\ 0 \end{pmatrix} + \begin{pmatrix}-7.5 \\ -12.99\end{pmatrix} = \begin{pmatrix}-43.5 \\ -12.99\end{pmatrix} \, \mathrm{km}\) Now, we can find the total displacement magnitude: \(|\Delta d|= \sqrt{(-43.5)^2 + (-12.99)^2} \approx 45.20 \, \mathrm{km}\) Since we already have the total time, \(t = \frac{1}{2} \, \mathrm{h}\), we can find the average velocity: \(\mathrm{Average \, Velocity} = \frac{|\Delta d|}{t} = \frac{45.20}{0.5} = 90.4 \, \mathrm{km/h}\) The average speed for the trip is \(102\, \mathrm{km/h}\), and the average velocity for the trip is \(90.4\, \mathrm{km/h}\).