I need help, worth 40 Points! A plane flies due north (90° from east) with a velocity of 100 km/h for 3 hours. During this time, a steady wind blows southeast at 30 km/h at an angle of 315° from due east. After 3 hours, where will the plane’s position be relative to its starting point?

*Please try to fill the data table.

Split up the two given velocity vectors into horizontal and vertical components.

Plane:

p = (100 km/h) (cos(90°) i + sin(90°) j ) = (100 km/h) j

Its direction is given, 90° relative to east.

Traveling for 3 hours with this velocity results in a displacement of

p * (3 h) = (300 km/h) (3h) j = (900 km) j

meaning that this velocity contributes to a north-facing displacement of 900 km.

Wind:

w = (30 km/h) (cos(315°) i + sin(315°) j ) ≈ (21.2 km/h) i + (-21.2 km/h) j

Its direction is also given, 315° relative to east.

After 3 h, this velocity contributes to a displacement of

w * (3h) ≈ (63.6 km) i + (-63.6 km) j

meaning if the plane had no velocity of its own but somehow stayed in the air, the wind would have pushed it about 63.6 km south and 63.6 km east, which translates to a net displacement of

√((63.6 km)² + (-63.6 km)²) = 90 km

due southeast.

Putting everything together, your table displacement table should read:

.                       |   wind   |   plane

velocity           |   30      |   100       ...   km/h

direction         |   90°     |   315°      ...   relative to east

x                      |   63.6   |   0           ...   km

y                      |  -63.6   |   900      ...   km

The resultant vector is the sum of these, r = p + w :

r ≈ (21.2 km/h) i + (78.8 km/h) j

After 3 h, the resultant displacement is

r * (3h) ≈ (63.6 km) i + (236.4 km) j

with magnitude

√((63.6 km)² + (236.4 km)²) ≈ 244.8 km

and direction θ such that

tan(θ) ≈ (236.4 km) / (63.6 km)

θ ≈ arctan(3.714) ≈ 74.9°

So the resultant table should read

xnet             |   63.6 km

ynet             |   236.4 km

magnitude   |   244.8 km

theta            |   74.9° relative to east