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Bond [772]
3 years ago
13

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.
Physics
2 answers:
tankabanditka [31]3 years ago
5 0

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

Andrei [34K]3 years ago
5 0

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°

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