Question

An airplane pilot sets a compass course due west and maintains an airspeed of 210 km/h. After flying for a time of 0.490 h, she finds herself over a town a distance 121 km west and a distance 20 km south of her starting point.Find the magnitude of the wind velocity.

Answer 1

Answer:

$V_{A/E}=55.23km/h\\or\\V_{A/E}=15.342m/s$

Explanation:

The velocity of plane relative to earth is given by:

$V_{P/E}=\frac{d}{t}\\V_{P/E}=\frac{[-121i-20j]km}{0.490h} \\V_{P/E}=[-247i-41j]km/h$

As the from given data.The velocity of plane relative to air is:

$V_{P/A}=[-210i]km/h$

According to relative motion of velocity of the air relative to earth given by:

$V_{A/E}=V_{P/E}-V_{P/A} \\V_{A/E}=[(-247i-41j)km/h]-(-210i)km/h\\V_{A/E}=[-37i-41j]km/h$

The magnitude is given as:

$V_{A/E}=\sqrt{(37)^{2}+(41)^{2} }\\ V_{A/E}=55.23km/h\\or\\V_{A/E}=15.342m/s$

Answer 2

Answer:

55km/h , 47.87° south west or 227.87°

Explanation:

If there is no wind then the plane would be at a distance

d=(210 km/h)(0.49h) = 102.9 km west of the starting point.

if the wind is blown then then the plane is 121 km - 102.9 km = 18.1 km west and 20 km south in the time 0.49 h.

Velocity of the plane in west direction=(18.1 km) / 0.49 h = 36.9 km/h

velocity of the plane in south direction=(20 km) / 0.49 h = 40.8 km/h

Now the wind velocity is

$\sqrt{(36.9^2 + 40.8^2}$

= 55.01 km/h

The direction is

θ =tan⁻¹ (40.8 / 36.9)

θ = 47.87° south west or 227.87°