Answer to Question #138157 in Physics for Reneth

Question

Kims home is 10km away from school, she took displacement from the GPS tracker of their car, her notes are A=6km,53 deg. SE, B=2km,63 deg. SW, C=1.5km,51deg.NW,D=5km S, E=3km,F=5km N, G=1.5km,44deg.SW,H=2km,63deg.NW
What is the value of 0 in shortcut path?
What is the value of the shortcut path along the x axis in km

Answer 1

All these vectors have the following coordinates.

$$\vec{OA} = (6\cos 53\degree, 6\sin 53\degree ) \approx(3.61, 4.79 )\\ \vec{AB} = (-2\sin 63\degree, -2\cos 63\degree ) \approx(-1.78, -0.91 )\\ \vec{BC} = (-1.5\cos 51\degree, 1.5\sin 51\degree ) \approx(-0.94, 1.17 )\\ \vec{CD} = (0, -5 ) \\ \vec{DE} = (0, 3 )\\ \vec{EF} = (0, 5 ) \\ \vec{FG} = (-1.5\cos 44\degree, -1.5\sin 44\degree ) \approx(-1.08, -1.04 )\\ \vec{GH} = (-2\sin 63\degree, 2\cos 63\degree ) \approx(-1.78, 0.91 )$$

The shortcut will be:

$$\vec{OH} = \vec{OA} + \vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{EF} + \vec{FG} + \vec{GH}=\\ = (-1.98, 7.92)$$

The length of the shortcut pass is $8.16\space km$. The value of the shortcut path along the x axis is $1.98 \space km$.

Answer. 8.16 km, 1.68 km.