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:

Step1: Find the x and y components of each displacement

First displacement: \( d_1 = 3\,\text{km} \) along x - axis. So, \( x_1 = 3\,\text{km} \), \( y_1 = 0\,\text{km} \)

Second displacement: \( d_2 = 2\,\text{km} \) at \( \theta = 45^\circ \) above x - axis.
The x - component of the second displacement: \( x_2=d_2\cos\theta=2\cos45^\circ = 2\times\frac{\sqrt{2}}{2}=\sqrt{2}\approx1.414\,\text{km} \)
The y - component of the second displacement: \( y_2 = d_2\sin\theta=2\sin45^\circ=2\times\frac{\sqrt{2}}{2}=\sqrt{2}\approx1.414\,\text{km} \)

Step2: Find the total x and y components of the resultant displacement

Total x - component: \( x = x_1 + x_2=3+\sqrt{2}\approx3 + 1.414 = 4.414\,\text{km} \)
Total y - component: \( y = y_1 + y_2=0+\sqrt{2}\approx1.414\,\text{km} \)

Step3: Find the distance from the origin (magnitude of resultant displacement)

The magnitude of the resultant displacement \( r \) is given by the Pythagorean theorem: \( r=\sqrt{x^{2}+y^{2}} \)
Substitute \( x = 3+\sqrt{2} \) and \( y=\sqrt{2} \)
\( r=\sqrt{(3 + \sqrt{2})^{2}+(\sqrt{2})^{2}}=\sqrt{9 + 6\sqrt{2}+2 + 2}=\sqrt{13 + 6\sqrt{2}} \)
Calculate the numerical value: \( 6\sqrt{2}\approx6\times1.414 = 8.484 \), \( 13+8.484 = 21.484 \), \( \sqrt{21.484}\approx4.635\,\text{km} \)

Step4: Find the angle above the x - axis

The angle \( \theta \) is given by \( \tan\theta=\frac{y}{x} \)
Substitute \( y = \sqrt{2} \) and \( x = 3+\sqrt{2} \)
\( \tan\theta=\frac{\sqrt{2}}{3+\sqrt{2}} \)
Rationalize the denominator: \( \frac{\sqrt{2}(3 - \sqrt{2})}{(3+\sqrt{2})(3 - \sqrt{2})}=\frac{3\sqrt{2}-2}{9 - 2}=\frac{3\sqrt{2}-2}{7} \)
Calculate the numerator: \( 3\sqrt{2}-2\approx3\times1.414 - 2=4.242 - 2 = 2.242 \)
\( \tan\theta=\frac{2.242}{7}\approx0.320 \)
\( \theta=\arctan(0.320)\approx17.7^\circ \)

Answer:

The distance from the origin is approximately \( \boldsymbol{4.64\,\text{km}} \) (rounded to two decimal places) and the angle above the positive x - axis is approximately \( \boldsymbol{17.7^\circ} \) (rounded to one decimal place)