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Sapling Learning macmilan leaming One day you go hiking at a nearby nature prese

ID: 1863533 • Letter: S

Question



Sapling Learning macmilan leaming One day you go hiking at a nearby nature preserve. At first, you follow the straight, clearly marked trails. From your starting point, you travel 2.00 miles down the 1st trail. Then you turn to your left by 30.0 to follow a 2nd trail for 1.10 miles. Next, you turn to your right by 160 and follow a 3rd trail for 1.80 miles. At this point you are getting very tired and would like to get back as quickly as possible, but all of the available trails seem to lead you deeper into the woods. You would like to take a shortcut directly through the woods (ignoring the trails). How far to your right should you turn, and how far do you have to walk, to go directly back to your starting point? Feel free to use the provided vector drawing board to help visualize your work. 0.25 miles Number Note: The vector drawing Turn 161 to the right, uide and a reality check, but may not give you the answers with sufficient precision. It is preferable to calculate the magnitude and direction by hand. then walk 1.7 miles Previous Give Up & View Solution Check Answer NextExit Hint

Explanation / Answer

Keep track of position as you move along. Let's say the first path points in the direction of the +x-axis.

End of first path:

Ax = 2.00 mi

End of second path:

Ax = 2.00 + 1.10cos30.0° mi

Ax = 2.953 mi

Ay = 1.10sin30.0° mi

Ay = 0.55 mi

End of third path:

Ax = 2.953 + 1.8cos(30.0° - 160°) mi

Ax = 1.796 mi

Ay = 0.55 + 1.8sin(30.0° - 160°) mi

Ay = -0.829 mi

Reference angle to to help you get back to where you started:

tan(th) = Ay / Ax

tan(th) = (-0.829) / (1.796)

(th) = -24.77° (24.77° clockwise from +x-axis)

But we need to go the other direction:

180 - 24.77° = +155.2°

And you're facing 30.0° - 160° = -130° (or 360° - 130° = +230°) at the end of the third path, so you'll need to turn

230° - 155.2° = 74.77° to the right

And walk a distance:

A = sqrt[(Ax)² + (Ay)²]

A = sqrt[(1.796 mi)² + (-0.829 mi)²]

A = 1.98 mi

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