Two airplanes leave an airport at the same
time. The velocity of the first airplane is
710 m/h at a heading of 27.5
◦
. The velocity
of the second is 650 m/h at a heading of 167◦
.
How far apart are they after 3.3 h?
Answer in units of m
figure the distance each has traveled, and use the law of cosines to get the final separation.
Start by drawing a diagram, so you know what's what.
To find the distance between the two airplanes after 3.3 hours, we need to use the concepts of velocity and displacement.
Step 1: Convert the velocities into vector form.
The velocity of the first airplane is 710 m/h at a heading of 27.5 degrees. In vector form, this can be written as (710*cos(27.5), 710*sin(27.5)).
The velocity of the second airplane is 650 m/h at a heading of 167 degrees. In vector form, this can be written as (650*cos(167), 650*sin(167)).
Step 2: Calculate the displacements of the airplanes after 3.3 hours.
The displacement of an object can be calculated by multiplying its velocity by the time traveled. So, for each airplane, we need to multiply its velocity by 3.3 hours.
For the first airplane, the displacement would be (3.3*710*cos(27.5), 3.3*710*sin(27.5)).
For the second airplane, the displacement would be (3.3*650*cos(167), 3.3*650*sin(167)).
Step 3: Find the distance between the two displacements.
To find the distance between the two displacements, we can use the Pythagorean theorem. The distance can be calculated using the formula: distance = sqrt((x2 - x1)^2 + (y2 - y1)^2).
So, distance = sqrt((x2 - x1)^2 + (y2 - y1)^2), where (x1, y1) is the displacement of the first airplane and (x2, y2) is the displacement of the second airplane.
Step 4: Calculate the distance.
Substitute the values of the displacements into the distance formula and calculate the answer in units of meters.
distance = sqrt((3.3*650*cos(167) - 3.3*710*cos(27.5))^2 + (3.3*650*sin(167) - 3.3*710*sin(27.5))^2)
Evaluate this expression to find the distance between the two airplanes after 3.3 hours. The answer will be in units of meters.