Two airplanes leave an airport at the same
time. The velocity of the first airplane is
690 m/h at a heading of 67.5
The velocity
of the second is 620 m/h at a heading of 182◦
How far apart are they after 3.5 h?
Answer in units of m.
All angles are measured CW from +y-axis.
d1 = 690[67.5o] * 3.5 = 2415m[67.5].
d2 = 620[182o] * 3.5 = 2170m[182o].
D = d1 - d2 = 2415[67.5] - 2170[182].
X = 2415*sin67.5 - 2170*sin182 = 2231 m.
Y = 2415*Cos67.5 - 2170*Cos182 = 3093 m.
D = sqrt(X^2 + Y^2).
Well, if the airplanes are flying at different angles, we can't really use their velocities directly to find the distance between them. We'll have to use a little bit of trigonometry here, so brace yourself for some calculations!
First, let's break down the velocities into their horizontal and vertical components:
For the first airplane:
Horizontal component = 690 m/h * cos(67.5°)
Vertical component = 690 m/h * sin(67.5°)
For the second airplane:
Horizontal component = 620 m/h * cos(182°)
Vertical component = 620 m/h * sin(182°)
Now, let's find how far each airplane has traveled after 3.5 hours. We can calculate that by multiplying the respective horizontal and vertical components by 3.5:
For the first airplane:
Horizontal distance = (690 m/h * cos(67.5°)) * 3.5
Vertical distance = (690 m/h * sin(67.5°)) * 3.5
For the second airplane:
Horizontal distance = (620 m/h * cos(182°)) * 3.5
Vertical distance = (620 m/h * sin(182°)) * 3.5
But we're not done yet! To find the distance between the two airplanes, we need to calculate the horizontal and vertical differences between their positions. So, subtract the corresponding horizontal and vertical distances:
Horizontal difference = (690 m/h * cos(67.5°)) * 3.5 - (620 m/h * cos(182°)) * 3.5
Vertical difference = (690 m/h * sin(67.5°)) * 3.5 - (620 m/h * sin(182°)) * 3.5
Finally, we can use the Pythagorean theorem to find the distance between the two airplanes:
Distance = sqrt((Horizontal difference)^2 + (Vertical difference)^2)
Whew! That's a lot of math! I hope you're ready to punch those numbers in and find the answer!
To find the distance between the two airplanes after 3.5 hours, we can use the concept of relative velocity.
Step 1: Convert the velocities from m/h to m/s.
- Velocity of the first airplane: 690 m/h = (690 * 1000) / 3600 m/s = 191.67 m/s
- Velocity of the second airplane: 620 m/h = (620 * 1000) / 3600 m/s = 172.22 m/s
Step 2: Calculate the components of the velocities in the x and y directions.
- For the first airplane:
- Velocity in the x direction: Vx1 = 191.67 m/s * cos(67.5°)
- Velocity in the y direction: Vy1 = 191.67 m/s * sin(67.5°)
- For the second airplane:
- Velocity in the x direction: Vx2 = 172.22 m/s * cos(182°)
- Velocity in the y direction: Vy2 = 172.22 m/s * sin(182°)
Step 3: Calculate the combined x and y distances for each airplane after 3.5 hours.
- For the first airplane:
- Distance in the x direction: Dx1 = Vx1 * time = 191.67 m/s * 3.5 h
- Distance in the y direction: Dy1 = Vy1 * time = 191.67 m/s * 3.5 h
- For the second airplane:
- Distance in the x direction: Dx2 = Vx2 * time = 172.22 m/s * 3.5 h
- Distance in the y direction: Dy2 = Vy2 * time = 172.22 m/s * 3.5 h
Step 4: Calculate the total distance between the two airplanes using the distance formula.
- Distance between the airplanes:
- Distance = sqrt((Dx2 - Dx1)^2 + (Dy2 - Dy1)^2)
Now, let's calculate the values step-by-step:
First, calculate the components of the velocities for each airplane:
- For the first airplane:
- Vx1 = 191.67 m/s * cos(67.5°) ≈ 77.94 m/s
- Vy1 = 191.67 m/s * sin(67.5°) ≈ 175.01 m/s
- For the second airplane:
- Vx2 = 172.22 m/s * cos(182°) ≈ -172.22 m/s (negative value because of the opposite direction)
- Vy2 = 172.22 m/s * sin(182°) ≈ 1.97 m/s
Next, calculate the distances traveled by each airplane after 3.5 hours:
- For the first airplane:
- Dx1 = Vx1 * time = 77.94 m/s * 3.5 h ≈ 272.99 m
- Dy1 = Vy1 * time = 175.01 m/s * 3.5 h ≈ 612.54 m
- For the second airplane:
- Dx2 = Vx2 * time = -172.22 m/s * 3.5 h ≈ -602.77 m (negative value because of the opposite direction)
- Dy2 = Vy2 * time = 1.97 m/s * 3.5 h ≈ 6.90 m
Finally, calculate the total distance between the two airplanes using the distance formula:
- Distance = sqrt((Dx2 - Dx1)^2 + (Dy2 - Dy1)^2)
= sqrt((-602.77 m - 272.99 m)^2 + (6.90 m - 612.54 m)^2)
≈ sqrt((-875.76 m)^2 + (-605.64 m)^2)
≈ sqrt(767801.68 m^2 + 367389.64 m^2)
≈ sqrt(1133341.32 m^2)
≈ 1064.91 m
Therefore, the two airplanes are approximately 1064.91 meters apart after 3.5 hours.
To find the distance between the two airplanes after 3.5 hours, we can use the formula for displacement. The formula is given by:
displacement = velocity * time
First, let's calculate the displacements of each airplane after 3.5 hours.
For the first airplane:
displacement_1 = velocity_1 * time
= 690 m/h * 3.5 h
For the second airplane:
displacement_2 = velocity_2 * time
= 620 m/h * 3.5 h
Now, we have the displacements of each airplane. However, we need to consider the directions in which they are moving.
To find the final distance between the airplanes, we will use the Pythagorean theorem because the displacements of the two airplanes form a right triangle. The sides of the triangle will be the absolute values of the displacements.
distance = √((displacement_1)^2 + (displacement_2)^2)
Let's substitute the displacements and calculate the distance:
distance = √((690 m/h * 3.5 h)^2 + (620 m/h * 3.5 h)^2)
distance = √((2415 m)^2 + (2170 m)^2)
distance = √(5,842,225 m^2 + 4,724,900 m^2)
distance = √10,567,125 m^2
distance ≈ 3250.75 m
Therefore, after 3.5 hours, the two airplanes are approximately 3250.75 meters apart.