Two Dimensional Vectors, level 13
In this video we will go over how to solve distance and bearing problems. In the fields
of surveying and navigation, the bearing of a line (L) is the acute angle formed by a
fixed north-south line and the line (L). The notation used to designate the bearing of
a line begins with N or S (for north or south), followed by the number of degrees in the angle,
and ends with E or W (for East or West). All bearings are measured in a horizontal plane.
For example the following bearings would be denoted as follows, the bearing of point B
from A would be pronounced as 40 degrees east of north, notice that we start on the line
that faces north and then move 40 degrees towards the east. The second bearing would
be denoted as follows, and would be pronounced as 65 degrees west of north, once again notice
that we start on the line facing north and move towards the west 65 degrees and stop.
The third bearing would be denoted as follows, and it would be pronounced as 70 degrees east
of South, and finally the last bearing would be denoted as follows and it will be pronounced
as 20 degrees west of south. If you are dealing with a problem that involves air navigation,
the bearing of an air plane will usually be measured in degrees and clockwise from north.
This can be confusing because it goes in the opposite direction of the standard angle convention
learned in your math courses. For example if these same bearings represented the direction
of an air plane they would be denoted as follows, the first vector would have a bearing of 40
degrees, the second vector would have a bearing of 295 degrees, and the third vector would have
a bearing of 110 degrees, and the last vector would have a bearing of 200 degrees. Alright
let’s go over a couple of examples.
A boat travels 72 miles on a course of bearing 27 degrees east of north and then changes
its course to travel 37 miles at 55 degrees east of north. How far north and how far east
has the boat traveled on this 109-mile trip? Alright we can go ahead and solve this problem
by using vectors to represent each of the 2 bearings with the magnitude of each vector
representing the distance traveled. We can figure out how far north and how far east
the boat traveled by finding the resultant vector and using its components to determine
how far the boat traveled.
Alright for the first 72 mile trip we go ahead and start at the origin and draw a vector
that starts north and moves towards the east at an angle of 27 degrees as follows. This
bearing represents the first 72 mile trip. Let’s denote this vector as vector V.
Now for the next trip we need to draw the cardinal directions on the tip of vector
V and use it to draw the next bearing, in this case we once again start at the direction
facing north and move 55 degrees towards the east as follows. Let’s label this vector
as vector U.
Now that we have both vectors we can go ahead and find the components of each vector and
then add the vectors component wise in order to find the components of the resultant vector.
Looking at the diagram we see that we can use either the 27 degree angle or 63 degree
angle to find the components of vector V and we can use either 55 degrees or 35 degrees
to find the components of vector U, it really does not matter which angle you use as long
as you apply the trigonometric ratios correctly, using 63 degrees for vector V we obtain the
following expression for the components, and using 35 degrees for vector U we obtain the
following for the components.
Now it is just a matter of finding the resultant vector by adding vector V and U component
wise, doing that we obtain 63 miles for the x-component and 85 miles for the y-component.
So the boat traveled 63 miles towards the east and 85 miles towards the north. Alright
let’s go over the next example. A person is riding in a hot air balloon. For
the first hour and a half the wind current is a constant 22.0 miles per hour in the direction
37.5 degrees east of north. Then the wind current changes to 18.5 miles per hour and
heads the balloon in the direction 52.5 degrees east of south. If this continuous for another
2 hours, how far is the balloon from its starting point? What is the bearing of the balloon
from its starting point? Similar to the previous problem let’s
first draw a diagram to get a visual of the movements of this air balloon. Let’s start
with the first bearing we know that this bearing will be 37.5 degrees east from north. Now
we need to figure out the distance that this balloon travels in this particular direction.
We know from the problem that the balloon will travel at 22 miles per hour for an hour
and a half. We can easily find the distance traveled by using the classical distance equals
rate times time formula. Computing the distance for this first trip we obtain 33 miles. Let’s
go ahead and label this vector as vector V. Then the balloon changes direction and will
now move at a bearing of 52.5 degrees east of south so we draw the cardinal directions
on the head of the first vector. From here we start south and move towards the east at
an angle of 52.5 degrees. Next we need to find the distance traveled on this second
trip, this time around the balloon is moving at 18.5 miles per hour for 2 hours, once again
we make use of the formula, obtaining 37 miles. Let’s call this vector, vector U.
At this point it is just a matter of adding both vectors V and U component wise. Let’s
first find the components of each vector. Using the 37.5 degree angle we obtain the
following expression for the components of vector V, and using the 52.5 degree angle
we obtain the following expression for the components of vector U, notice that the y-component
is negative since it is pointing towards the negative y-axis. Next we go ahead
and add Vector V and vector U component wise, doing that we obtain the following.
Finally we go ahead and find the magnitude of the resultant vector which is approximately
equal to 49.6 miles. We can easily find the direction by using the components of the resultant
vector by using inverse tangent, doing that we obtain 4.23 degrees. This angle represents
the angle that the vector makes with the positive x-axis we need to take 90 degrees and subtract
this angle to obtain the requested bearing, doing that we obtain 85.8 degrees east from
north. Alright let’s go over the final example. An airplane is traveling at a fixed altitude
with a negligible wind factor. The airplane is traveling at a speed of 500 miles per hour
with a bearing of 330 degrees. As the airplane reaches a certain point, it encounters wind
with a velocity of 70 miles per hour in the direction 45 degrees east of north. What are
the resultant speed (ground speed) and direction of the airplane?
Alright in this problem the path and speed of an airplane is being altered due to wind,
this is similar to the way in which a boat's path and speed can be altered by water current.
Because of this wind we need to find the true course or track of the plane.
This is nothing more than the direction of the resultant velocity of the plane and wind.
We also want to find the ground speed of the plane this would be equal to the magnitude
of the resultant velocity. Note that speed refers to the magnitude of the velocity vector;
in physics these two quantities are different but related to each other with speed being
a scalar and velocity being a vector.
Ok like always let’s start by drawing a diagram, since this is an air navigation problem
the bearing will be measured from the north clockwise. So a bearing of 330 degrees would
be represented as follows. This is also equivalent to 120 degrees measured from the positive
x-axis. Next we go ahead and draw the vector representing the wind velocity in this case
it will point 45 degrees east of north. Now it is just a matter of finding the components
of each velocity, for the components of the velocity of the plane I will use the 120 degree
angle this will take care of the negative and positive signs on the components automatically,
finding an expression for the components we obtain the following, and doing the same for
the wind velocity we obtain the following. Next we go ahead and add the vectors component
wise, doing that we obtain the following for the resultant velocity.
The final step is to find the magnitude and direction, in this case the magnitude will
be approximately equal to 522.5 miles per hour, and the direction will be equal to 67.4
degrees after using inverse tangent but we need to adjust this angle, adjusting the angle
and using the standard conventions used in air navigation we obtain 337.4 degrees.
So the new speed of the airplane would be 522.5 miles per hour at a bearing of 337.4
degrees. Alright and this ends the two dimensional vector series. In our next series of videos
we will go over three dimensional vectors.