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? 3.77 km 4.41 km 14.21 km 19.46 km clear all

Explanation:

Step1: Identify vectors and angle

Let the first - displacement vector $\vec{A}=3.65\ km$ (east) and the second - displacement vector $\vec{B} = 0.94\ km$ (40° south). The angle between the two vectors $\theta=40^{\circ}$.

Step2: Apply the law of cosines

The magnitude of the resultant displacement $R$ is given by the law of cosines $R=\sqrt{A^{2}+B^{2}-2AB\cos(180 - \theta)}$. Since $\cos(180-\theta)=-\cos\theta$, the formula becomes $R=\sqrt{A^{2}+B^{2}+2AB\cos\theta}$. Substitute $A = 3.65$, $B = 0.94$, and $\theta = 40^{\circ}$ (so $\cos\theta=\cos40^{\circ}\approx0.766$).
\[

$$\begin{align*} R&=\sqrt{(3.65)^{2}+(0.94)^{2}+2\times3.65\times0.94\times0.766}\\ &=\sqrt{13.3225 + 0.8836+2\times3.65\times0.94\times0.766}\\ &=\sqrt{13.3225 + 0.8836+(7.3\times0.94\times0.766)}\\ &=\sqrt{13.3225 + 0.8836+(6.862\times0.766)}\\ &=\sqrt{13.3225 + 0.8836 + 5.2563}\\ &=\sqrt{19.4624}\\ &\approx3.77 \end{align*}$$

\]

Answer:

3.77 km