Question

The pilot of a small plane finds that the airport where he intended to land is fogged in. He flies 62.6 mi west to another airport to find that conditions there are too icy for him to land. He flies 28.9 mi at 15.0° east of south and is finally able to land at the third airport. Assume north to be +y and east to be +x. How many extra miles beyond his original flight plan has he flown?

Answer 1

Answer:

$d = 61.8 miles$

Explanation:

first pilot moved 62.6 miles west towards another airport

so we have

$d_1 = -62.6 \hat i$

now at that airport it fly again 28.9 miles at 15 East of South

$d_2 = 28.9 sin15 \hat i - 28.9 cos15 \hat j$

so we have

$d_2 = 7.47 \hat i - 27.9 \hat j$

now net displacement of the plane is given as

$d = d_1 + d_2$

$d = -62.6 \hat i + 7.47 \hat i - 27.9 \hat j$

$d = -55.13 \hat i - 27.9 \hat j$

magnitude of the displacement is given as

$d = \sqrt{55.13^2 + 27.9^2}$

$d = 61.8 miles$