Two airplanes leave an airport at the same
time. The velocity of the first airplane is
730 m/h at a heading of 44.6 degrees.
The velocity
of the second is 610 m/h at a heading of 100 degrees.
How far apart are they after 2.3 h?
Answer in units of m.
Thank you, but it did actually say meters haha
Well, it sounds like these airplanes are on quite the adventure! Let's see how far apart they are after 2.3 hours.
To calculate the distance between them, we can use the concept of vector addition. Think of the velocities of the airplanes as vectors with magnitude and direction. The distance between them can be found using the formula:
distance = sqrt((change_in_x)^2 + (change_in_y)^2)
Now let's break this down:
For the first airplane, the change in x (horizontal direction) after 2.3 hours can be calculated by multiplying its velocity (730 m/h) by the cosine of its heading angle (44.6 degrees):
change_in_x_first_airplane = 730 m/h * cos(44.6 degrees)
Similarly, the change in y (vertical direction) after 2.3 hours can be calculated by multiplying its velocity (730 m/h) by the sine of its heading angle (44.6 degrees):
change_in_y_first_airplane = 730 m/h * sin(44.6 degrees)
For the second airplane, we'll do the same calculations using its velocity (610 m/h) and heading angle (100 degrees):
change_in_x_second_airplane = 610 m/h * cos(100 degrees)
change_in_y_second_airplane = 610 m/h * sin(100 degrees)
Now, to find the total change in x and y for both airplanes, we'll add these individual changes:
total_change_in_x = change_in_x_first_airplane + change_in_x_second_airplane
total_change_in_y = change_in_y_first_airplane + change_in_y_second_airplane
Finally, we can use the distance formula mentioned earlier to find the distance between the two airplanes after 2.3 hours:
distance = sqrt(total_change_in_x^2 + total_change_in_y^2)
Now that we have all the pieces, let's calculate this distance and get an answer in meters!
To find the distance between two airplanes after a given time, we can use the concept of relative velocity. This involves considering the velocities of both airplanes and their respective headings.
1. First, we can break down the velocities of the airplanes into their horizontal and vertical components. This can be done using trigonometry, specifically the sine and cosine functions.
For the first airplane:
Horizontal component = velocity * cos(theta)
= 730 m/h * cos(44.6 degrees)
Vertical component = velocity * sin(theta)
= 730 m/h * sin(44.6 degrees)
For the second airplane:
Horizontal component = velocity * cos(theta)
= 610 m/h * cos(100 degrees)
Vertical component = velocity * sin(theta)
= 610 m/h * sin(100 degrees)
2. Next, we calculate the change in position (distance) of each airplane after 2.3 hours. This is done by multiplying their respective velocities (horizontal and vertical components) by the time:
Distance for the first airplane:
Horizontal distance = (730 m/h * cos(44.6 degrees)) * 2.3 h
Vertical distance = (730 m/h * sin(44.6 degrees)) * 2.3 h
Distance for the second airplane:
Horizontal distance = (610 m/h * cos(100 degrees)) * 2.3 h
Vertical distance = (610 m/h * sin(100 degrees)) * 2.3 h
3. Finally, we can find the total distance between the two airplanes by using the Pythagorean theorem. This theorem states that the square of the hypotenuse (total distance) is equal to the sum of the squares of the two shorter sides (horizontal and vertical distances) in a right triangle.
Total distance = sqrt((Horizontal distance)^2 + (Vertical distance)^2)
Plug in the values from the previous calculations and solve for the total distance. The resulting value will be the answer to the question, expressed in units of meters.
remember the law of cosines?
the angle between the two headings is 55.4°
So, figuring distance = speed * time, we have your distance apart (z) after 2.3 hours is
z^2 = (730*2.3)^2 + (610*2.3)^2 - 2(730*2.3)(610*2.3)cos55.4°
z = 1453 m
I suspect you meant mi (miles), not m (meters), since no plane could get off the ground at 600 meters/hr.
Of course, in that case, they are flying faster than any commercial airliners in use today!