2. an airplane flying undergoes two consecutive displacements. the first is 75.0 km at 30.0° west of north…

Explanation:

Step1: Resolve the first - displacement vector

The first displacement $\vec{d}1 = 75.0$ km at $30.0^{\circ}$ west of north. In the x - y coordinate system (north - south along y - axis and east - west along x - axis), the x - component of $\vec{d}1$ is $d{1x}=-75\sin30^{\circ}$ km and the y - component is $d{1y}=75\cos30^{\circ}$ km. $d_{1x}=-75\times\frac{1}{2}=- 37.5$ km, $d_{1y}=75\times\frac{\sqrt{3}}{2}\approx64.95$ km.

Step2: Resolve the second - displacement vector

The second displacement $\vec{d}2 = 155$ km at $60.0^{\circ}$ east of north. The x - component of $\vec{d}2$ is $d{2x}=155\sin60^{\circ}$ km and the y - component is $d{2y}=155\cos60^{\circ}$ km. $d_{2x}=155\times\frac{\sqrt{3}}{2}\approx134.23$ km, $d_{2y}=155\times\frac{1}{2}=77.5$ km.

Step3: Calculate the total x - component of the resultant displacement

$d_x=d_{1x}+d_{2x}=-37.5 + 134.23=96.73$ km.

Step4: Calculate the total y - component of the resultant displacement

$d_y=d_{1y}+d_{2y}=64.95 + 77.5 = 142.45$ km.

Step5: Calculate the magnitude of the resultant displacement

The magnitude of the resultant displacement $\vec{d}$ is given by $d=\sqrt{d_x^{2}+d_y^{2}}$. $d=\sqrt{(96.73)^{2}+(142.45)^{2}}=\sqrt{9356.69 + 20291.00}=\sqrt{29647.69}\approx172$ km.

Step6: Calculate the direction of the resultant displacement

The direction $\theta$ (measured from the north) is given by $\tan\theta=\frac{d_x}{d_y}$. $\tan\theta=\frac{96.73}{142.45}\approx0.679$. $\theta=\arctan(0.679)\approx34.1^{\circ}$ east of north.

Answer:

The magnitude of the resultant displacement is approximately $172$ km at an angle of $34.1^{\circ}$ east of north.