Question

1. (6pts) beginning from the origin, you walk 3 km along the positive x - axis and then walk 2 km at an angle of 45° above the positive x - axis. what is your distance from the origin (3pts) and, if a friend at the origin wanted to walk straight towards you, at what angle above the positive x - axis would your friend go (3pts)?

Explanation:

Part 1: Distance from the origin

Step 1: Find x and y components of each displacement

First, the first displacement is 3 km along the x - axis. So its x - component \(x_1 = 3\space km\) and \(y_1=0\space km\).

The second displacement is 2 km at an angle of \(45^{\circ}\) above the x - axis. Using the formulas for components: \(x_2 = 2\cos(45^{\circ})\) and \(y_2=2\sin(45^{\circ})\). Since \(\cos(45^{\circ})=\sin(45^{\circ})=\frac{\sqrt{2}}{2}\approx0.707\), we have \(x_2 = 2\times\frac{\sqrt{2}}{2}=\sqrt{2}\space km\approx1.414\space km\) and \(y_2 = 2\times\frac{\sqrt{2}}{2}=\sqrt{2}\space km\approx1.414\space km\).

Step 2: Find the total x and y components

The total x - component of the displacement from the origin is \(x=x_1 + x_2=3+\sqrt{2}\space km\approx3 + 1.414=4.414\space km\)

The total y - component of the displacement from the origin is \(y=y_1 + y_2=0+\sqrt{2}\space km\approx1.414\space km\)

Step 3: Calculate the distance from the origin

Using the distance formula \(d=\sqrt{x^{2}+y^{2}}\), we substitute \(x = 3+\sqrt{2}\) and \(y=\sqrt{2}\)

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Part 2: Angle your friend should walk

Step 1: Recall the formula for the angle

The angle \(\theta\) above the x - axis is given by \(\tan(\theta)=\frac{y}{x}\), where \(x = 3+\sqrt{2}\) and \(y=\sqrt{2}\)

Step 2: Calculate the tangent of the angle

\(\tan(\theta)=\frac{\sqrt{2}}{3+\sqrt{2}}\). Rationalize the denominator:

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Step 3: Find the angle

\(\theta=\arctan(0.320)\approx17.7^{\circ}\)

Answer:

(Distance from origin): \(\boldsymbol{\approx4.64\space km}\) (rounded to two decimal places)