Two airplanes leave an airport at the same
time. The velocity of the first airplane is
650 m/h at a heading of 40◦
. The velocity of
the second is 560 m/h at a heading of 108◦
.
How far apart are they after 1.5 h?
Answer in units of m.
A, 650 * 1.5 = 975
B, 560 * 1.5 = 840
angle between them = 108 - 40 = 68
draw parallelogram bottom 975, left 840, top 975, right 840, angle 68 at bottom left
c^2 = 840^2 + 975^2 - 840*975 cos 68
= 705600 + 950625 - 819000 *.3746
= 349,427
so diagonal c = 1,161
To find the distance between two airplanes after 1.5 hours, we can use the concept of relative velocity.
Step 1: Determine the x and y components of velocity for each airplane.
For the first airplane:
Velocity = 650 m/h
Heading = 40°
To find the x-component:
Vx1 = Velocity * cos(Heading)
= 650 * cos(40°)
To find the y-component:
Vy1 = Velocity * sin(Heading)
= 650 * sin(40°)
For the second airplane:
Velocity = 560 m/h
Heading = 108°
To find the x-component:
Vx2 = Velocity * cos(Heading)
= 560 * cos(108°)
To find the y-component:
Vy2 = Velocity * sin(Heading)
= 560 * sin(108°)
Step 2: Calculate the relative velocity components.
The x-component of the relative velocity (Vx_rel) is the difference between Vx1 and Vx2:
Vx_rel = Vx1 - Vx2
The y-component of the relative velocity (Vy_rel) is the difference between Vy1 and Vy2:
Vy_rel = Vy1 - Vy2
Step 3: Find the distance traveled by the two airplanes after 1.5 hours.
The distance traveled by an airplane is given by:
Distance = Velocity * Time
Distance1 = Velocity1 * Time
= 650 * 1.5
= 975 m
Distance2 = Velocity2 * Time
= 560 * 1.5
= 840 m
Step 4: Calculate the final distance between the two airplanes.
Using the Pythagorean theorem, the final distance (D) between the two airplanes is given by:
D = sqrt((Distance1 + Vx_rel)^2 + (Distance2 + Vy_rel)^2)
D = sqrt((975 + Vx_rel)^2 + (840 + Vy_rel)^2)
Now, we substitute the values of Vx_rel and Vy_rel, and calculate the final distance.