Two small planes leave the Abbotsford airport at the same time. The first flies at 225km/h at a heading 320°, while the second flies at 190km/h at the heading of 70°. How far apart are they after 2 hours?
320 is 40 degrees west of north
north speed= +225 cos 40
east speed = -225 sin 40
70 is 70 degrees east of north
north speed= 190 cos 70
east speed = 190 sin 70
difference in north speed
=225cos40-190 sin70
=-6.18 km/h
difference in east speed
=-225sin40-190sin 70
=-323.17 km/h
magnitude of difference in speed
= sqrt (6.18^2+323^2)
= 323 km/h speed apart
* 2 hours
= 646 km
The angle between their flight paths is 110°
I sketched the triangle and used the cosine law.
x^2 = 450^2 + 380^2 - 2(4500)(380)cos110°
= 463870.889
x = √463870.889 = appr 681 km
or using vectors and the actual angles
let v = (450cos320, 450sin320) = (344.72,-289.25)
let u = (380cos70, 380sin70) = (129.97, 357.08)
let w be the vector joining the end of v to the start of w
then v + w - u = 0
w = u - v = (129.97, 357.08) - (344.72,-289.25)
= (-214.75, 646.34)
|w| = √(214.75^2 + 646.34^2) = appr 681 km , as before
To find out how far apart the two planes are after 2 hours, we can use the concept of vectors and vector addition.
First, let's find the displacement vector for each plane after 2 hours.
For the first plane:
Distance = Speed * Time
Distance = 225 km/h * 2 hours = 450 km
To obtain the displacement vector, we need to consider both the magnitude (450 km) and direction (heading 320°). The displacement vector for the first plane can be represented as:
d₁ = 450 km at 320°
For the second plane:
Distance = Speed * Time
Distance = 190 km/h * 2 hours = 380 km
Similarly, we can represent the displacement vector for the second plane as:
d₂ = 380 km at 70°
To find the distance between the two planes, we need to add these two displacement vectors. We can break down each displacement vector into its horizontal (x) and vertical (y) components using trigonometry.
For the first plane:
x₁ = 450 km * cos(320°)
y₁ = 450 km * sin(320°)
For the second plane:
x₂ = 380 km * cos(70°)
y₂ = 380 km * sin(70°)
Using these formulas, we can calculate the x and y components:
x₁ = 450 km * cos(320°) = -286.16 km
y₁ = 450 km * sin(320°) = -307.80 km
x₂ = 380 km * cos(70°) = 187.92 km
y₂ = 380 km * sin(70°) = 353.55 km
Now, we can calculate the difference between the x-components and the y-components:
∆x = x₂ - x₁
∆y = y₂ - y₁
∆x = 187.92 km - (-286.16 km) = 474.08 km
∆y = 353.55 km - (-307.80 km) = 661.35 km
The distance between the two planes after 2 hours can then be found using the Pythagorean theorem:
Distance = √(∆x² + ∆y²)
Distance = √(474.08 km)² + (661.35 km)²)
Distance ≈ √(224,733.86 km² + 437,484.22 km²)
Distance ≈ √662,218.08 km²
Distance ≈ 814.15 km
Therefore, the two planes are approximately 814.15 km apart after 2 hours.