Question

multiple choice question
you travel 58.3 km in the direction of 15.0° north of east, then travel 77.3 km directly south. what is the magnitude of your net displacement? (round to three significant figures.)
108 km
83.9 km
136 km
571 km

Explanation:

Step1: Calculate the x - component of the first displacement

The first displacement \(d_1 = 58.3\) km at an angle \(\theta=15.0^{\circ}\) north of east.
The x - component \(d_{1x}=d_1\cos\theta\)
\(d_{1x}=58.3\cos(15.0^{\circ})\)
\(d_{1x}=58.3\times0.966 = 56.3\) km

Step2: Calculate the y - component of the first displacement

The y - component \(d_{1y}=d_1\sin\theta\)
\(d_{1y}=58.3\sin(15.0^{\circ})\)
\(d_{1y}=58.3\times0.259 = 15.1\) km

Step3: Calculate the y - component of the second displacement

The second displacement \(d_2 = 77.3\) km south (so \(d_{2y}=- 77.3\) km, \(d_{2x} = 0\) km)

Step4: Calculate the net x - component

\(D_x=d_{1x}+d_{2x}=56.3 + 0=56.3\) km

Step5: Calculate the net y - component

\(D_y=d_{1y}+d_{2y}=15.1-77.3=-62.2\) km

Step6: Calculate the magnitude of the net displacement

Using the Pythagorean theorem \(D=\sqrt{D_x^{2}+D_y^{2}}\)
\(D=\sqrt{(56.3)^{2}+(-62.2)^{2}}\)
\(D=\sqrt{3169.69 + 3868.84}\)
\(D=\sqrt{7038.53}\)
\(D = 83.9\) km

Answer:

83.9 km