2 You are standing on the ground at the origin of a coordinate system.
3 An airplane flies over you with constant velocity parallel to the $x$
4 axis at a fixed height of $7.60\E{3}\U{m}$. At time $t=0$, the
5 airplane is directly above you so that the vector leading from you to
6 it is $\vect{P}_0=7.60\E{3}\jhat\U{m}$. At $t=30.0\U{s}$, the
7 position vector leading from you to the airplane is
8 $\vect{P}_{30}=(8.04\E{3}\ihat+7.60\E{3}\jhat)\U{m}$ as suggested in
9 Figure~P3.43. Determine the magnitude and orientation of the
10 airplane's position vector at $t=45.0\U{s}$.
20 draw((-dx,h)--(d+dx,h), dashed);
21 Surface s = Surface((-dx,0), (d+dx,0));
24 Vector A = Vector((0,0), mag=h, dir=90,
25 L=Label("$\vect{R}_0$", position=EndPoint, align=W));
27 Vector B = Vector((0,0), mag=length((d,h)), dir=degrees((d,h)),
28 L=Label("$\vect{R}_{30}$", position=EndPoint, align=RightSide));
35 The airplane is flying at a constant velocity, so we'll compute that
38 \vect{v} &= \frac{\vect{P}_{30} - \vect{P}_0}{t_{30} - t_{0}}
39 = \frac{(8.04\E{3}\ihat+7.60\E{3}\jhat)\U{m} - 7.60\E{3}\jhat\U{m}}
40 {30.0\U{s} - 0\U{s}} \\
41 &= \frac{8.04\E{3}\ihat}{30.0}\U{m/s}
44 Which is, as claimed in the problem text, parallel to the $x$ axis.
46 The position at $t=45.0\U{s}$ is therefore
48 \vect{P}_{45} &= \vect{P}_0 + \vect{v}(t_{45}-t_0)
49 = 7.60\E{3}\jhat\U{m} + 268\ihat\U{m/s} \cdot (45.0\U{s} - 0\U{s}) \\
50 &= (12.1\E{3}\ihat + 7.60\E{3}\jhat)\U{m}
51 = (12.1\ihat + 7.60\jhat)\U{km} \\
52 |\vect{P}_{45}| &= \sqrt{\vect{P}_{45,x}^2 + \vect{P}_{45,y}^2}
53 = \sqrt{12.1^2 + 7.60^2}\U{km}
55 \theta_{45} &= \arctan\p({\frac{\vect{P}_{45,y}}{\vect{P}_{45,x}}})
56 = \arctan\p({\frac{7.60}{12.1}})