Question

1. a cross - county ski trail begins at a parking lot and heads due west for 14.0 km. it then goes for 9.5 km on a heading of 40° north of west. at this point, the trail takes a bearing 65° for 6.0 km and ends at a ski chalet. determine the displacement of the chalet relative to the parking lot.

Explanation:

Step1: Resolve each vector into components

Let's assume the west - east direction as the \(x\) - axis (negative \(x\) for west) and the north - south direction as the \(y\) - axis (positive \(y\) for north).

• For the first vector \(\vec{A}\) ( \(A = 14.0\space km\) due west):

\(A_x=- 14.0\space km\), \(A_y = 0\space km\)

• For the second vector \(\vec{B}\) (\(B = 9.5\space km\) at \(40^{\circ}\) north of west):

\(B_x=-B\cos40^{\circ}=-9.5\cos40^{\circ}\approx - 9.5\times0.766=-7.277\space km\)
\(B_y = B\sin40^{\circ}=9.5\sin40^{\circ}\approx9.5\times0.643 = 6.109\space km\)

• For the third vector \(\vec{C}\) (\(C = 6.0\space km\) at \(65^{\circ}\) (assuming from the positive \(x\) - axis, if it's a bearing, we need to adjust. If the bearing is \(65^{\circ}\) from the north - south meridian, we assume it's \(65^{\circ}\) east of north. Let's assume it's \(65^{\circ}\) from the positive \(x\) - axis for simplicity. If it's a standard bearing (e.g., from the north, clockwise), we need to re - calculate. Here we assume it's \(65^{\circ}\) from the positive \(x\) - axis)

\(C_x=C\cos65^{\circ}=6.0\cos65^{\circ}\approx6.0\times0.423 = 2.538\space km\)
\(C_y=C\sin65^{\circ}=6.0\sin65^{\circ}\approx6.0\times0.906=5.436\space km\)

Step2: Sum the \(x\) and \(y\) components

• Sum of \(x\) - components:

\(R_x=A_x + B_x+C_x=-14.0-7.277 + 2.538=-18.739\space km\)

• Sum of \(y\) - components:

\(R_y=A_y + B_y+C_y=0 + 6.109+5.436 = 11.545\space km\)

Step3: Calculate the magnitude and direction of the resultant vector

• Magnitude of the resultant vector \(R\):

\(R=\sqrt{R_x^{2}+R_y^{2}}=\sqrt{(- 18.739)^{2}+(11.545)^{2}}=\sqrt{351.15+133.38}=\sqrt{484.53}\approx22.0\space km\)

• Direction \(\theta\):

\(\tan\theta=\frac{|R_y|}{|R_x|}=\frac{11.545}{18.739}\approx0.616\)
\(\theta=\arctan(0.616)\approx31.6^{\circ}\) north of west

Answer:

The displacement of the chalet relative to the parking lot is approximately \(22.0\space km\) at an angle of \(31.6^{\circ}\) north of west.