Relative Velocity

Question

An aircraft leaves A to fly to B which is 100 km due North of A. The pilot sets a course due North but after 15 minutes he realizes that, owing to a wind blowing from due East, the plane is at a point C, where C is 65 km from A and 16 km West of A. Find

 

 

(a) the speed , in km/h , of the wind,

 

(b) the speed , in km/h , of the aircraft in still air,

 

 (c) the course the pilot should have set from A in order to have arrived directly at B.

 

Find also the course the pilot should set from C in order to fly directly to B and , in this case, the time, to the nearest minute , taken for the journey from A to B .

Answer 1

a) The plane flies north, but the wind moves the plane 16 kilometers west of its destination in 15 minutes. The speed of the wind is:

 

v = s/t = (16 km)/(.25 hours) = 64 km/he

 

b) The plane with the wind traveled 65 miles in 15 minutes, but it was 16 miles west of its destination (which is north). This means the distance it traveled north can be found by using the Pythagorean Theorem.

 

16^2 + x^2 = 65^2

256 + x^2 = 4225

x^2 = 3969

x = 63 km

 

The plane traveled 63 km north in 15 minutes which it would traveled in still air since it wouldn't move west without the wind. The speed of the plane in still air would be:

 

v = s/t = (63 km)/(.25 hours) = 252 km/hr

 

c) To find the direct spot the plane should set for the arrive directly at B, we should a ratio of the two right triangles that would represent each course. 

 

63/16 = 100/x

 

63x = 1600

x = 25.4 km

 

The plane should travel to a point 25.4 km west of destination B from A. This means that the plane shouldn't travel directly north, but at an west of true north. That angle would be:

 

Tan O = 25.4/100 = .254

 

O = 14.25 degrees

 

The plane should travel at an angle 14.25 degrees west of true north.