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.

Explanation:

The velocity of plane relative to earth is given by:

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

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

The magnitude is given as:

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

= 55.01 km/h

The direction is

θ =tan⁻¹ (40.8 / 36.9)

θ = 47.87° south west or 227.87°

well, you would divide 135 by 33 to get your answer. then that is how many seconds it would take. boardman should fly about 132 m or 4.090909090909091 seconds.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s