Question

A spelunker is surveying a cave. She follows a passage 140 m straight west, then 210 m in a direction 45 east of south, and then 280 m at 30 east of north. After a fourth unmeasured displacement, she nds herself back where she started. Determine the magnitude and direction of the fourth displacement.

Answer 1

Answer:

301.96 m with direction of 10.4 degrees north of west

Explanation:

Let i and j be the unit vector of east and north directions, respectively. So if she moving 140m straight west then her distance vector would be

$\vec{s_1} = -140\hat{i}$

When she walks 210 m in a direction 45 east of south, her displacement vector is:

$\vec{s_2} = 210sin45^0\hat{i} - 210cos45^0\hat{j}$

$\vec{s_2} = 210/\sqrt{2}\hat{i} - 210/\sqrt{2}\hat{j}$

When she walks 280 m in a direction 30 east of north, her displacement vector is:

$\vec{s_3} = 280sin30^0\hat{i} + 280cos30^0\hat{j}$

$\vec{s_3} = 140\hat{i} + 140\sqrt{3}\hat{j}$

Then she walk another passage to be back to the origin

$\vec{s_1} + \vec{s_2} + \vec{s_3} + \vec{s_4} = 0\hat{i} + 0\hat{j}$

$140\hat{i} - 210/\sqrt{2}\hat{i} + 210/\sqrt{2}\hat{j} - 140\hat{i} - 140\sqrt{3}\hat{j} = x_4\hat{i} + y_4\hat{j}$

where x4 and y4 are the displacement she made in the east and north direction in her 4th displacement:

$x_4 = 140 - 210\sqrt{2} - 140 = -210\sqrt{2} \approx -297$

$y_4 = 210/\sqrt{2} - 140\sqrt{3} \approx 54.5$

So her last displacement vector is

$\vec{s_4} = -297 \hat{i} + 54.5 \hat{j}$

This vector would have a displacement and direction of:

$s_4 = \sqrt{y_4^2 + x_4^2} = \sqrt{54.5^2 + -297^2} = \sqrt{2970.25 + 88209} = \sqrt{91179.25} = 301.96 m$

$tan\alpha= \frac{y_4}{x_4} = \frac{54.5}{297} = 0.18$

$\alpha = tan^{-1}0.18 = 0.18 rad \approx 10.4^o$ north of west