Imagine a trip where you drive along an east-west highway at $80.0 \mathrm{km} / \mathrm{h}$ for 45.0 min and then you turn onto a highway that runs \(38.0^{\circ}\) north of east and travel at \(60.0 \mathrm{km} / \mathrm{h}\) for 30.0 min. (a) What is your average velocity for the trip? (b) What is your average velocity on the return trip when you head the opposite way and drive \(38.0^{\circ}\) south of west at \(60.0 \mathrm{km} / \mathrm{h}\) for the first \(30.0 \mathrm{min}\) and then west at \(80.0 \mathrm{km} / \mathrm{h}\) for the last \(45.0 \mathrm{min} ?\)

First, we need to find the individual displacements for each part of the trip. Displacement is the product of velocity and time. Since all movement occurs in the horizontal (east-west) or diagonal (north-east or south-west) directions, we can simply find the distances traveled in each part of the trip and deal with the directions in the next step. For the forward trip: Displacement during the east-west travel: \(80.0 \mathrm{km/h} \times 0.75\mathrm{h} = 60.0\mathrm{km}\) Displacement during the north-east travel: \(60.0 \mathrm{km/h} \times 0.5\mathrm{h} = 30.0\mathrm{km}\) For the return trip: Displacement during the south-west travel: \(60.0 \mathrm{km/h} \times 0.5\mathrm{h} = 30.0\mathrm{km}\) Displacement during the west travel: \(80.0 \mathrm{km/h} \times 0.75\mathrm{h} = 60.0\mathrm{km}\)

Next, we need to convert the individual displacements in the diagonal directions into vectors with horizontal and vertical components. This will allow us to easily add them to find the total displacement. For the forward trip: East-west displacement: \(60.0\mathrm{km} \,\hat{i}\) (east direction) The north-east displacement can be split into components as follows: \(x = 30.0\mathrm{km} \cos (38.0^{\circ}) \approx 23.5\mathrm{km} \,\hat{i}\) \(y = 30.0\mathrm{km} \sin (38.0^{\circ}) \approx 18.6\mathrm{km} \,\hat{j}\) For the return trip: West displacement: \(60.0\mathrm{km} \,(-\hat{i})\) (west direction) The south-west displacement can be split into components as follows: \(x = -30.0\mathrm{km} \cos (38.0^{\circ}) \approx -23.5\mathrm{km} \,\hat{i}\) \(y = -30.0\mathrm{km} \sin (38.0^{\circ}) \approx -18.6\mathrm{km} \,\hat{j}\)

Now that we have all the displacements as vectors with components, we can add them to find the total displacements for the forward and return trips. For the forward trip: \(\vec{D}_{forward} = 60.0\hat{i} + 23.5\hat{i} + 18.6\hat{j} = (83.5\hat{i} + 18.6\hat{j})\,\mathrm{km}\) For the return trip: \(\vec{D}_{return} = -60.0\hat{i} - 23.5\hat{i} - 18.6\hat{j} = (-83.5\hat{i} - 18.6\hat{j})\,\mathrm{km}\)

To find the average velocities for both trips, we need the total time. The total time is the sum of individual trip times: Total time for forward trip and return trip: \(0.75\mathrm{h} + 0.5\mathrm{h} = 1.25\mathrm{h}\)

Finally, we can calculate the average velocities by dividing the total displacements by the total time: For the forward trip: \(\vec{V}_{avg,forward} = \frac{\vec{D}_{forward}}{t} = \frac{83.5\hat{i} + 18.6\hat{j}}{1.25} \approx (66.8\hat{i} + 14.9\hat{j})\,\mathrm{km/h}\) For the return trip: \(\vec{V}_{avg,return} = \frac{\vec{D}_{return}}{t} = \frac{-83.5\hat{i} - 18.6\hat{j}}{1.25} \approx (-66.8\hat{i} - 14.9\hat{j})\,\mathrm{km/h}\) So the average velocities for the forward trip and return trip are approximately \((66.8\hat{i} + 14.9\hat{j})\,\mathrm{km/h}\) and \((-66.8\hat{i} - 14.9\hat{j})\,\mathrm{km/h}\), respectively.