College Physics 3.18 – Driving in a straight line vs east and north distances

You drive 7.50 km in a straight line in a direction 15° east of north.

(a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.)

(b) Show that you still arrive at the same point if the east and north legs are reversed in order.


Part A

Consider the following figure:

North and East Components of the given displacement
North and East Components of the given displacement

The east distance is the component in the horizontal direction.

D_E=7.50\:km\:\cdot sin\:\left(15^{\circ} \right)


The north distance is the vertical component

D_E=7.50\:km\cdot cos\left(15^{\circ} \right)


Part B


Based from the figure, we can easily see that the order is reversible in the addition of vectors. We say that D_E+D_N=D_N+D_E