At some point as you are hiking near a lake, you determine that your campsite is 1.50 km away from you in the direction 30.0° E of N. However, to get back to your campsite, you will need to walk around the lake. You set off due north and walk for 600 m. You then turn in the direction 20.0 W of N and walk an addol1.20 km, before turning and walking directly to your campsite. How far and in what direction was the last leg of your hike?

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aristocles aristocles · Answer 1 · 2019-09-23T03:00:51+02:00

Answer:

[tex]r_2 = 976.65 m[/tex]

Direction is 19 degree South of East

Explanation:

Let say initial position is our reference

so we will have campsite position given as

[tex]r = 1.50 km[/tex] at 20 degree E of N

now we will have

[tex]r = 1500 sin20\hat i + 1500 cos20\hat j[/tex]

[tex]r = 513 \hat i + 1409.5\hat j[/tex]

now our displacement to walk around is given as

[tex]d_1 = 600 \hat j[/tex]

then we move 20 degree W of N and move 1200 m

so we will have

[tex]d_2 = 1200 sin20(-\hat i) + 1200cos20\hat j[/tex]

so our final position is given as

[tex]r_1 = d_1 + d_2[/tex]

[tex]r_1 = 600\hat j - 410.4 \hat i + 1127.6\hat j[/tex]

[tex]r_1 = -410.4 \hat i + 1727.6\hat j[/tex]

now we know that

[tex]r_1 + r_2 = r[/tex]

so final leg of the displacement is given as

[tex]r_2 = r - r_1[/tex]

[tex]r_2 = (513 \hat i + 1409.5\hat j) - (-410.4 \hat i + 1727.6\hat j)[/tex]

[tex]r_2 = 923.4\hat i - 318.1 \hat i[/tex]

so magnitude is given as

[tex]r_2 = \sqrt{923.4^2 + 318.1^2}[/tex]

[tex]r_2 = 976.65 m[/tex]

direction is given as

[tex]\theta = tan^{-1}\frac{y}{x}[/tex]

[tex]\theta = tan^{-1}\frac{-318.1}{923.4}[/tex]

[tex]\theta = -19 degree[/tex]

so it is 19 degree South of East