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 ○ r² = 70² - 40² ○ r² = 40² + 70² ○ r² = 40² + 70² - 2(40)(70)cos(135) ○ r² = 40² + 70² - 2(40)(70)cos(45)

Explanation:

Step1: Identify vectors and angle

Let the first - displacement vector $\vec{A}=40\mathrm{m}$ east and the second - displacement vector $\vec{B}=70\mathrm{m}$ southeast. The angle between the two vectors $\theta = 45^{\circ}$.

Step2: Apply the law of cosines

The magnitude of the resultant displacement $R$ of two vectors $\vec{A}$ and $\vec{B}$ is given by the law of cosines: $R^{2}=A^{2}+B^{2}-2AB\cos(180^{\circ}-\theta)$. Since the angle between the two vectors in the formula for the law of cosines is the angle between them when placed tail - to - tail, and the angle between the east - direction and the southeast - direction is $45^{\circ}$, the angle used in the law - of - cosines formula is $135^{\circ}$. Here $A = 40$, $B = 70$, so $R^{2}=40^{2}+70^{2}-2(40)(70)\cos(135^{\circ})$.

Answer:

The correct formula for the magnitude of the total displacement is $R^{2}=40^{2}+70^{2}-2(40)(70)\cos(135^{\circ})$. So the answer is the option with $R^{2}=40^{2}+70^{2}-2(40)(70)\cos(135^{\circ})$.