Question

1. leland runs 3.65 km east, then turns 40° south and runs another 0.94 km. what is the magnitude of his displacement? clear all 4.41 km 3.77 km 1.41 km 19.46 km

Explanation:

Step1: Resolve the second - displacement into components

The first displacement $\vec{d}_1 = 3.65\ km$ east. The second displacement $\vec{d}_2=0.94\ km$ at $40^{\circ}$ south. The x - component of the second displacement is $d_{2x}=0.94\cos(40^{\circ})\ km$ and the y - component is $d_{2y}=- 0.94\sin(40^{\circ})\ km$.
$d_{2x}=0.94\times\cos(40^{\circ})\approx0.94\times0.766 = 0.720\ km$
$d_{2y}=-0.94\times\sin(40^{\circ})\approx - 0.94\times0.643=-0.604\ km$
The total x - component of the displacement $\vec{d}$ is $d_x=d_{1}+d_{2x}=3.65 + 0.720=4.37\ km$
The total y - component of the displacement $\vec{d}$ is $d_y=d_{2y}=- 0.604\ km$

Step2: Calculate the magnitude of the total displacement

The magnitude of the displacement vector $\vec{d}$ is given by $d=\sqrt{d_x^{2}+d_y^{2}}$.
$d=\sqrt{(4.37)^{2}+(-0.604)^{2}}=\sqrt{19.0969 + 0.3648}=\sqrt{19.4617}\approx4.41\ km$

Answer:

4.41 km