### Video Transcript

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.