Answer to Question #241859 in Mechanics | Relativity for Kim

Question

Axel leaves his house, drives 23 km due North, then turns onto a street and continues in a direction 60° N of E for 40 km and finally turns onto the highway due East for 20 km. What is his total displacement from his house?

Answer 1

The movement of Axel can be traced as shown in the figures below;

Taking O as the initial point, first displacement, $d_1=(23km)y$ (O to A)

Second displacement, $d_2 =(40cos60x+40sin60y)$ (A to B)

Third displacement, $d_3=(20km)x$ (B to C)

Total displacement, $d=d_1+d_2+d_3=23y+40cos60x+40sin60y+20x$

$=(20+40cos60)x+(23+40sin60)y=(40x+57.6y)km$

Magnitude, $d=\sqrt{40^2+57.6^2}=70Km$, $angle \space \theta =tan^{-1}(57.6/40)=55^o$

So d=70km, $55^o$ North of East