A plane flies \(410 \mathrm{mi}\) east from city \(A\) to city \(B\) in \(45 \mathrm{~min}\) and then \(820 \mathrm{mi}\) south from city \(B\) to city \(C\) in 1 h \(30 \mathrm{~min}\). (a) What are the magnitude and direction of the displacement vector that represents the total trip? What are \((b)\) the average velocity vector and \((c)\) the average speed for the trip?

The total displacement is calculated by vector addition. Because these vectors are perpendicular (east and south are perpendicular directions), they form a right-angled triangle. The displacement to the east can be represented by a vector \([410,0]\) and the displacement to the south by a vector \([0,-820]\). Adding these vectors gives \([410,-820]\). Using Pythagoras' theorem (\( \sqrt{x^2 + y^2} \)), we get the magnitude of displacement as \( \sqrt{410^2 + (-820)^2} = 900 \) miles. To find the direction, calculate the inverse tangent of the absolute value of the vertical displacement divided by the horizontal displacement. So \( \theta = \tan^{-1}(\frac{-820}{410}) = -63.43° \). Bearing in mind that south of east is a negative angle, our final angle is \(63.43°\) south of east.

The average velocity is calculated by dividing the displacement vector by the total time for the journey. First, convert all the times into the same units to calculate the total travel time, which is \(45 min + 1.5h = 1.75 h\). The average velocity vector is the total displacement vector \([410, -820]\) divided by the total time \(1.75h\), thus \([234.3, -468.6]\) miles/hr. In other words, the velocity is \(900/1.75 = 514.29\) miles/hr. The direction remains \(63.43°\) south of east as the average velocity vector and displacement vector always have the same direction.

The average speed is calculated by dividing total distance travelled by the total time for the journey. In this case, the total distance travelled is \(410 + 820 = 1230 \) miles, and the total time is still \(1.75 h\). Dividing the total distance by the total time, the average speed works out to \(1230 / 1.75 = 702.86\) miles/hr.