### Video Transcript

In this video, we’re going to learn
about displacement vectors. What they are, and how to calculate
them both graphically and algebraically. To get started, imagine that you’re
taking part in a neighborhood scavenger hunt. Where, given a list of clues, you
move from one location to another until you get to the last clue and the last
location where the game ends. Each contestant starts at the base
of the tree in the southwest corner of town. And clue by clue, you navigate the
course until finally at the end, you receive the very last clue and are able to find
the last spot. Just as you’re completing the
course, a friend of yours tells you that they’re starting out at the course
beginning. If you wanted to give them a
direction that they could go straight to the end point where you currently are, what
would you tell them? It’s displacement vectors that will
help us understand this idea more clearly.

The first thing we might want to
understand with this topic is, what is a displacement vector? And to figure that out, let’s look
at one on a graph. Let’s imagine that this 𝑥𝑦 pair
of axes is put onto a table top. And then an ant sits on that
tabletop in an initial given position. Based on this frame of reference,
the starting position of the ant, which we can call 𝑝 sub one, is given by a vector
from the origin to the ant’s location. Say that the ant then begins to
move. And after travelling a path for
some time ends up in a second position that we can call 𝑝 sub two. If we were to take these two
position vectors and subtract the first from the second, then the result would be a
vector that we can call 𝑑. This is the displacement of the
ant. Graphically, 𝑑 is a vector that
begins at the end of our first position vector and ends at the point or the tip of
our last position vector.

From this, we understand that
displacement vectors connect initial position vectors with final position
vectors. It shows the change in
position. In this example, we’ve found the
displacement vector 𝑑 using a graphical approach. We’ve connected the tip of the
initial position vector with the tip of the final position vector. It’s also possible to solve for 𝑑
using an algebraic approach, where 𝑝 sub two and 𝑝 sub one are given in their
component form. To get some practice with that
method of solving for displacement, let’s look at an example.

The F-35B is a short takeoff and
vertical landing fighter jet. An F-35B takes off vertically
upward to a height of 20.00 meters, while still facing horizontally. The fighter then follows a flight
path that is angled at 30.00 degrees above a line parallel to the ground, 20.00
meters vertically above ground. The fighter flies for a distance of
20.00 kilometers along this trajectory. What is the fighter’s final
displacement?

We can call the displacement of the
fighter after these maneuvers 𝑑, where we understand that 𝑑 is a vector with both
magnitude and direction. To begin on our solution, let’s
draw out a sketch of this F-35B’s motion. We’re told that the jet fighter
rises vertically a distance we’ve called ℎ of 20.00 meters. And then begins to move at an angle
of 30.00 degrees to the horizontal along a path of length 20.00 kilometers. The fighter’s displacement 𝑑 is
the vector that connects its initial location on the ground with its final location
at the end of its 20.00 kilometer of leg. It’s that vector that we want to
solve for based on this given information. We can start by orienting the
motion of our fighter to a set of coordinate axes. We’ll position a pair of 𝑥- and
𝑦-coordinate axes, so that their origin, the place where they cross, is the place
where our jet fighter began its journey. This means we can write our
displacement vector 𝑑 in the following form. A component, we’ll call it 𝑑 sub
𝑥, in the 𝑖-direction plus a component, we’ll call it 𝑑 sub 𝑦, in the
𝑗-direction.

So now, our task is to set out and
solve for 𝑑 sub 𝑥 and 𝑑 sub 𝑦. Looking first at 𝑑 sub 𝑥, we see
that that value, the 𝑥-component of the jet fighter’s motion, will be equal to 𝑙
times the cos of 30.00 degrees. We know that 𝑙 is given as 20.00
kilometers or 20.00 times 10 to the third meters. When we multiply that by the cos of
30.00 degrees, we find a value of 1.732 times 10 to the fourth meters. That’s 𝑑 sub 𝑥. Now we want to solve for 𝑑 sub 𝑦,
the vertical displacement of the jet fighter. Looking at our diagram, we see that
𝑑 sub 𝑦 is equal to ℎ plus 𝑙 times the sin of 30.00 degrees. Plugging in for ℎ and 𝑙, we once
again use a value of 20.00 times 10 to the third meters for 𝑙. And ℎ, we’re given as 20.00
meters. With these values entered on our
calculator, we find that 𝑑 sub 𝑦, to four significant figures, is 1.002 times 10
to the fourth meters. We insert this value for 𝑑 sub 𝑦
in our expression for displacement, 𝑑. And this gives us the final form
for the displacement of the jet fighter, 1.732 times 10 to the fourth meters in the
𝑖-direction and 1.002 times 10 to the fourth meters in the 𝑗-direction.

Now let’s look at another
displacement vector example, where we’ll find displacement magnitude as well as
direction.

A delivery man starts at the post
office, drives 40 kilometers north, then 20 kilometers west, then 60 kilometers
northeast, and finally, 50 kilometers north to stop for lunch. Find the magnitude of the delivery
man’s net displacement vector. Give your answer to a precision of
three significant figures. Find the angle north of east made
by the delivery man’s net displacement vector. Give your answer to a precision of
two significant figures.

In this two-part exercise, we want
to solve first for the magnitude of the net displacement vector of the delivery
man. We’ll call that magnitude of
𝑑. We also want to solve for the angle
north of east made by this net displacement vector. We’ll label that angle 𝜃. Let’s start out by drawing a sketch
of the delivery man’s motion. Beginning with our four compass
directions — north, south, east, and west — we can let the delivery man begin at the
intersection of these two axes. We’re told the first leg of the
delivery man’s journey is 40 kilometers north, then 20 kilometers west, then 60
kilometers northeast. And finally, 50 kilometers north to
the end point of the delivery man’s journey for that morning.

With the journey laid out
graphically like this, we can draw in our displacement vector 𝑑 that shows from the
start to the end of these four journey legs. We can see that this vector 𝑑 has
both a component to the north, we can call it 𝑑 sub 𝑁 in the northern
direction. And a component to the east, we can
call that 𝑑 sub 𝐸 in the eastern direction. In order to solve for the magnitude
of the displacement vector, we’ll want to solve for the displacement vector itself
first. And to do that, we’ll solve for 𝑑
sub 𝑁 and 𝑑 sub 𝐸. Starting with 𝑑 sub 𝑁, 𝑑 sub 𝑁
is equal to the sum of all the northern components of each leg of our journey. Noting that the third leg of our
journey, where we travel 60 kilometers northeast, involves motion 45 degrees above
the horizontal. We can write out 𝑑 sub 𝑁 as 40
kilometers plus 60 kilometers times the sin of 45 degrees plus 50 kilometers. When we add these three terms
together, we find a value, to four significant figures, of 132.4 kilometers.

Now we move on to solving for the
eastern component of displacement 𝑑 sub 𝐸. Looking at our diagram of the
delivery man’s motion, we can write the terms of 𝑑 sub 𝐸 as negative 20
kilometers, that’s when we’re moving to the west, plus 60 kilometers times the cos
of 45 degrees. Adding these terms together, to
four significant figures, they’re equal to 22.46 kilometers. We now have our displacement vector
𝑑, written out in terms of its northern and eastern components. To find the magnitude of that
displacement vector, we’ll take the square root of 𝑑 sub 𝑁 squared plus 𝑑 sub 𝐸
squared. Entering these values on our
calculator, to three significant figures, the magnitude of 𝑑 is 134 kilometers. That’s the net displacement of the
delivery man.

We next want to solve for 𝜃 which,
looking at our diagram, is the angle north of east at which the displacement vector
points. The tangent of that angle 𝜃 is
equal to 𝑑 sub 𝑁, the northern component of the displacement, divided by 𝑑 sub
𝐸, the eastern component. If we take the arc tangent of both
sides of the equation, plug in values for 𝑑 sub 𝑁 and 𝑑 sub 𝐸, and then enter
this expression on our calculator. We find that, to two significant
figures, 𝜃 is 80 degrees. That’s the angle north of east in
which the displacement vector points.

Let’s now summarize what we’ve
learnt about displacement vectors. Displacement vectors have a length
or magnitude equal to the shortest distance between a start point and an end
point. Adding vectors to get a net
displacement requires separating them by dimensions, 𝑥 and 𝑦. And displacement vectors, like all
vectors, have both a magnitude, or length, and a direction. So the next time you’re looking for
the shortest way to get from 𝐴 to 𝐵, consider using a displacement vector.