Question

1. an antelope trots 40m east, then turns at a 45° angle and trots 70m southeast. which choice gives the correct solution for the magnitude of the total displacement? clear all ○ (r^{2}=40^{2}+70^{2}-2(40)(70)cos(45)) ○ (r^{2}=40^{2}+70^{2}) ○ (r^{2}=70^{2}-40^{2}) ○ (r^{2}=40^{2}+70^{2}-2(40)(70)cos(135))

Explanation:

Step1: Identify vectors and angle

The antelope has two - displacement vectors. The first displacement $d_1 = 40m$ east and the second displacement $d_2=70m$ southeast. The angle between the two vectors is $\theta = 135^{\circ}$ (since the first is east - direction and the second is $45^{\circ}$ south of east).

Step2: Apply the law of cosines

The magnitude of the resultant displacement $R$ of two vectors $\vec{A}$ and $\vec{B}$ with an angle $\theta$ between them is given by the law of cosines: $R^{2}=A^{2}+B^{2}-2AB\cos\theta$. Here, $A = 40$, $B = 70$, and $\theta=135^{\circ}$, so $R^{2}=40^{2}+70^{2}-2(40)(70)\cos(135^{\circ})$.

Answer:

$R^{2}=40^{2}+70^{2}-2(40)(70)\cos(135^{\circ})$