Question

required information
a speedboat moves west at 108 km/h for 29.0 min. it then moves at 60.0° south of west at 90.0 km/h for 10.0 min.
what is the magnitude of average velocity for the trip?
\boxed{} km/h

Explanation:

Step1: Convert time to hours

First segment: $t_1 = \frac{29.0}{60} = 0.4833\ \text{h}$
Second segment: $t_2 = \frac{10.0}{60} = 0.1667\ \text{h}$
Total time: $t_{total} = 0.4833 + 0.1667 = 0.65\ \text{h}$

Step2: Calculate displacement of each segment

First segment (west, positive x-direction):
$d_1 = v_1 t_1 = 108 \times 0.4833 = 52.2\ \text{km}$ (x-component: $52.2\ \text{km}$, y-component: $0$)

Second segment ($60.0^\circ$ south of west):
x-component: $d_{2x} = v_2 t_2 \cos(60^\circ) = 90.0 \times 0.1667 \times 0.5 = 7.5015\ \text{km}$
y-component (south, negative y-direction): $d_{2y} = -v_2 t_2 \sin(60^\circ) = -90.0 \times 0.1667 \times \frac{\sqrt{3}}{2} \approx -12.99\ \text{km}$

Step3: Find total displacement components

Total x-displacement: $d_{total,x} = 52.2 + 7.5015 = 59.7015\ \text{km}$
Total y-displacement: $d_{total,y} = 0 - 12.99 = -12.99\ \text{km}$

Step4: Calculate total displacement magnitude

$$d_{total} = \sqrt{d_{total,x}^2 + d_{total,y}^2} = \sqrt{(59.7015)^2 + (-12.99)^2} \approx \sqrt{3564.27 + 168.74} = \sqrt{3733.01} \approx 61.1\ \text{km}$$

Step5: Compute average velocity magnitude

$v_{avg} = \frac{d_{total}}{t_{total}} = \frac{61.1}{0.65} \approx 94.0\ \text{km/h}$

Answer:

94.0 km/h