Tag Archives: north and east components of displacement

College Physics by Openstax Chapter 3 Problem 18


You drive 7.50 km in a straight line in a direction 15º 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.


Solution:

Part A

Consider the illustration shown.

Let DE be the east component of the distance, and DN be the north component of the distance.

\begin{align*}
D_E & = 7.50 \  \sin 15^\circ  \\
D_E & = 1.9411 \ \text{km} \\
D_E & = 1.94 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}
\begin{align*}
D_N & = 7.50 \  \cos 15^\circ  \\
D_N & = 7.2444\ \text{km} \\
D_N & = 7.24 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Part B

It can be obviously seen from the figure below that you still arrive at the same point if the east and north legs are reversed in order.


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College Physics by Openstax Chapter 3 Problem 3


Find the north and east components of the displacement for the hikers shown in Figure 3.50.

Figure 3.50

Solution:

Refer to Figure 3-3-A for the north and east components of the displacement s of the hikers.

Figure 3-3-A

Considering the right triangle formed. The north component is computed as

\begin{align*}
\text{s}_{\text{north}} & = \left( 5.00 \ \text{km}  \right)\sin 40^\circ  \\
\text{s}_{\text{north}} & = 3.21 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}

Using the same right triangle, the east component is computed as follows.

\begin{align*}
\text{s}_{\text{east}} & = \left( 5.00 \ \text{km}  \right)\cos 40^\circ  \\
\text{s}_{\text{east}} & = 3.83 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}

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