### Video Transcript

In this lesson, we’re going to
learn how to find the net result when combining multiple values of a kinematic
quantity, such as two displacements or two accelerations. When making such combinations,
vectors will be the best mathematical objects to represent our kinematic
quantities. So let’s recall how to work with
vectors and their components.

Vectors are mathematical objects
that have both a magnitude and a direction. We can conveniently represent
vectors as arrows, where the length of the arrow is the magnitude of the vector. And the direction the arrow is
pointing is the direction of the vector, in this case directly off to the right. Representing vectors as arrows
allows us to do certain types of vector calculations geometrically.

To do these calculations, it’s
helpful to be able to refer to the two ends of the vector unambiguously. So we’ll call this end the tail and
this end the head. We can easily see why vectors are
useful for representing kinematic quantities by considering a person pushing a box
across the floor. Let’s say our person exerts a force
of 10 newtons to push the box. Then 10 newtons is the magnitude of
the force. But just knowing this magnitude
doesn’t tell us if the box is being pushed to the left or the right.

Now, as we can clearly see from the
picture, the direction of the force is towards the right. So the full information about this
force includes both its magnitude and direction, which is why vectors are the
appropriate way to represent force. As we’ll see later, to describe any
of the kinetic quantities that we’re interested in, we’ll need both a magnitude and
a direction. So since vectors are the
appropriate choice for all of these, let’s continue to work out how to combine
vectors.

When we add two vectors together,
we’ll need to take into account that each one has a direction. So we’ll need a different procedure
than simply adding two numbers. By representing vectors as arrows,
we can actually perform this sum quite easily. For example, consider the sum of
these two vectors over here.

To actually perform this sum, we
need to align the tail of the second vector to the head of the first vector. We do this graphically by simply
drawing the second vector with its tail at the head of the first vector. The sum of these two vectors is now
the vector whose tail is at the tail of the first vector and whose head is at the
head of the second vector. And again, we find this
geometrically by simply drawing that arrow.

As we can see, the magnitude and
direction of this vector sum is different from the simple arithmetic sum of the
magnitudes and directions of our two original vectors. However, like addition of numbers,
vector addition is also commutative. This means that the order of
summation doesn’t matter. So drawing the blue vector at the
head of the orange vector gives the same result in sum vector as drawing the orange
vector at the head of the blue vector. Adding a pair of vectors this way
to get a single sum suggests another way that we can represent an individual
vector.

We’ll start off by drawing our
vector with its tail at the origin of some Cartesian axes. Now let’s imagine that we draw a
straight vertical line from the head of our vector to the horizontal axis. Then we can draw this vector along
the horizontal axis that goes exactly as far to the right as the head of our
original vector. We can also do the same for the
vertical axis.

Now, this vector along the vertical
axis goes exactly as far up as the head of our original vector. But now let’s look at this picture
carefully. The vertical vector would fit
exactly between the head of the horizontal vector and the head of the vertical
vector. The same is true of the horizontal
vector. It fits exactly between the head of
the vertical vector and the head of our original vector. But let’s look carefully at what
we’ve drawn. Whether we start with the vertical
vector or the horizontal vector, we have the tail of a second vector drawn at the
head of a first vector.

We then have a third vector whose
tail is the same as the tail of the first vector and whose head is the same as the
head of the second vector. But these are just the three
vectors involved in a vector sum. So our original magenta vector can
be represented as the sum of a horizontal orange vector and a vertical blue
vector. If we call our original vector 𝐕,
where we used a half arrow over a letter to represent that we’re talking about a
vector, then we call the horizontal and vertical vectors the components of 𝐕. And we usually use the symbol 𝐕
sub 𝑥 to represent the horizontal component and 𝐕 sub 𝑦 to represent the vertical
component. Symbolically, we’d write that 𝐕 is
the sum of 𝐕 𝑥 and 𝐕 𝑦.

This is a very powerful
representation because as we can see from our graph 𝐕 𝑥 and 𝐕 𝑦 are aligned with
the axes of our Cartesian system, which means they’re perpendicular. Qualitatively, this allows us to
independently talk about the horizontal and vertical components of 𝐕, even though
𝐕 itself is neither horizontal nor vertical. In technical language, we call 𝐕
𝑥 and 𝐕 𝑦 the projection of 𝐕 onto these axes. This just means that 𝐕 𝑥 and 𝐕
𝑦 are the components of 𝐕 that are parallel to the axes of our system.

Quantitatively, the fact that our
triangle formed by 𝐕 𝑥, 𝐕 𝑦, and 𝐕 is a right triangle means that we can easily
apply the Pythagorean theorem and do trigonometric calculations. Let’s let the symbol for a vector
without the half arrow on top represent the magnitude of that vector. Then we have, by the Pythagorean
theorem applied to this right triangle, that the magnitude of 𝑉 is equal to the
square root of the sum of the squares of the magnitude of 𝑉 𝑥 and the magnitude of
𝑉 𝑦. So this relates the magnitude of 𝑉
to 𝑉 𝑥 and 𝑉 𝑦.

Let’s see if we can also relate the
direction. Since our vector is drawn on some
Cartesian axes, we can represent the direction of the vector by the angle between
the vector and the horizontal axis. Let’s call this angle 𝜃. To relate 𝜃 to 𝑉 𝑥 and 𝑉 𝑦,
recall that the tangent of an acute angle in a right triangle is equal to the length
of the opposite side divided by the length of the adjacent side. In this triangle, 𝐕 𝑦 is opposite
𝜃 and 𝐕 𝑥 is adjacent to 𝜃. So the tan of 𝜃 is the magnitude
of 𝑉 𝑦 divided by the magnitude of 𝑉 𝑥.

Alternatively, we can now take the
inverse tangent of both sides of this equation. This gives us 𝜃, the direction of
𝑉, is equal to the inverse tangent of the magnitude of 𝑉 𝑦 divided by the
magnitude of 𝑉 𝑥. This gives us two equations, one
for the magnitude and one for the direction of the vector 𝐕 in terms of only the
magnitudes of its components. We don’t need to make any reference
to the direction of these components because we know that they’re perpendicular to
each other and parallel to the axes of our system.

Conversely, we can also find the
magnitudes of 𝑉 𝑥 and 𝑉 𝑦 from the magnitude of 𝑉 and its direction. Recall that the cosine of an acute
angle of a right triangle is the length of the adjacent side divided by the length
of the hypotenuse. This means that the length of 𝑉 𝑥
is equal to the length of 𝑉 times the cos of 𝜃. Similarly, to find the length of
the opposite side of an acute angle, we use the sine of the angle instead of the
cosine. So the magnitude of 𝑉 𝑦 is the
magnitude of 𝑉 times the sin of 𝜃.

These two equations now give us the
magnitude of the components of a vector in terms of the magnitude of the vector
itself and the vector’s direction. Note that, again, we’ve made no
reference to the direction of these vectors. This is because their directions
are predetermined. The vector 𝐕 𝑥 is always along
the horizontal direction, and the vector 𝐕 𝑦 is always along the vertical
direction.

We started by using our notion of
adding vectors to come to the idea of vector components. Now, we can do the reverse. We can use the equations we’ve
derived to improve our ability to add vectors.

Let’s now consider a sum of two
vectors, vector 𝐕 and vector 𝐔. We already know how to perform this
sum by drawing a vector from the tail of vector 𝐕 to the head of vector 𝐔. But before we do that step, let’s
draw the horizontal and vertical components of these two vectors. For vector 𝐕, the process is the
same as before. We draw a vertical line to the
horizontal axis, which tells us how long the horizontal component is. The vertical component is then the
vector between the head of the horizontal component and the vector 𝐕 itself.

And thus, just like before, we have
that 𝐕 is the sum of 𝐕 𝑥 and 𝐕 𝑦. We can do the same for 𝐔. But we have to account for the fact
that the tail of 𝐔 is not located at the origin of our system. This is not a problem. If we recall though, it really
defines the components of the vector is that the horizontal component is parallel to
the horizontal axis of the system. The vertical component is parallel
to the vertical axis of the system. And the sum of the components
equals the vector itself. In other words, the horizontal and
vertical components of a vector form the horizontal and vertical legs of the right
triangle that has the vector as the hypotenuse.

For the vector 𝐔, those two legs
are these dotted lines. So the horizontal leg is 𝐔 𝑥, and
the vertical leg is 𝐔 𝑦. So as we can see, we can still draw
the components of 𝐔 even though 𝐔 is not at the origin.

All right, let’s now draw the
vector that’s the sum of the vectors 𝐕 and 𝐔. Let’s call this vector 𝐖. We’re now looking for the
horizontal and vertical components of 𝐖. The components of 𝐖 will form the
legs of the right triangle with 𝐖 as the hypotenuse. On the horizontal axis, this will
be the vector 𝐕 𝑥 plus the vector represented by this orange dotted arrow. On the vertical axis, this will be
the vector represented by this blue dotted arrow plus the vector 𝐔 𝑦.

But now let’s look closely at these
two dotted vectors. The orange dotted vector is the
same as the vector 𝐔 𝑥. Similarly, the blue dotted vector
is the same as the vector 𝐕 𝑦. Both of these are true because the
two solid vectors and the two dotted vectors form a rectangle and opposite sides of
a rectangle are congruent.

Anyway, we can now easily see that
since the tail of 𝐔 𝑥 lined up to the head of 𝐕 𝑥 makes the vector 𝐖 𝑥, 𝐖 𝑥
equals 𝐕 𝑥 plus 𝐔 𝑥. By the same logic and using the
commutativity of vector addition, we have that 𝐖 𝑦 is equal to 𝐕 𝑦 plus 𝐔
𝑦. Remember though that both of our
equations for the components of 𝐖 came from our original relationship that 𝐖 is
equal to 𝐕 plus 𝐔. This leads us to a very important
conclusion about the sum of two vectors. The components of a vector that is
the sum of two other vectors are each individually equal to the sum of the
corresponding component of those two vectors.

So as we’ve already seen, the
horizontal component of 𝐖 is the sum of the horizontal components of 𝐕 and 𝐔. Now, we might reasonably wonder why
this helps us. It seems like we’ve just replaced
one vector equation with two vector equations. However, we’ve actually gained a
lot. We can’t evaluate the sum of 𝐕
plus 𝐔 without considering the direction of these vectors.

However, recall that every
horizontal component has the same direction. When we add two vectors with the
same direction, we simply add their magnitudes and keep the direction. This means that we can replace both
of our equations for the horizontal and vertical components of 𝐖 with their
nonvector equivalents. So the magnitude of 𝑊 𝑥 is equal
to the magnitude of 𝑉 𝑥 plus the magnitude of 𝑈 𝑥. And the same is true of the
magnitude of 𝑊 𝑦. It’s the magnitude of 𝑉 𝑦 plus
the magnitude of 𝑈 𝑦.

Again, the only reason we were able
to convert these vector equations into scalar equations is because all of the
vectors in each equation have the same direction. What this means is that by using
our previous equations that related the magnitude and direction of vectors to the
sides of their components, we can evaluate this vector sum as these two usually much
easier scalar sums.

Okay, this has been a whole lot of
abstract math. But it’s been worth it because we
can now apply this one set of ideas to any of our kinematic quantities. Each of the quantities we’re
interested in can be represented by a vector.

Displacement is the distance from a
reference point in a particular direction. So since displacement has a
magnitude and a direction, we can represent it as a vector. Similarly, velocity is the speed of
motion in a particular direction. So it, too, has a magnitude and a
direction and can be represented with a vector. Acceleration is the rate of change
of velocity. Since velocity has both magnitude
and direction, the change in velocity can either be a change in magnitude or
direction or both. So the acceleration has to carry
information about magnitude and direction. So it, too, can be represented as a
vector. Finally, force, by Newton’s second
law, is equal to mass times acceleration. We know acceleration is a
vector. So mass, a scalar, times a vector
is just another vector.

Furthermore, as we saw in our
example of a person pushing a box, in order to fully describe a force, we need to
supply both a size and a direction. Because each of these quantities
can be represented with a vector, we can also work with combinations of these
quantities the same way we’d work with combinations of vectors. Let’s see some examples of
this.

A length of road stretches north
for 10 kilometers from the edge of a town to where it intersects an eastward
road. A car is broken down on the
eastward road. And the displacement of the edge of
the town from the car has a magnitude of 24 kilometers. How far east of the intersection is
the car to the nearest kilometer?

To start answering this question,
let’s draw a diagram. We have a northbound road, an
eastbound road, a broken-down car, and a town. Here’s our northbound road, our
eastbound road, our car, and our town. We’re told that the intersection of
these roadways is 10 kilometers north of the town. We’re also told that the magnitude
of the displacement between the car and the town is 24 kilometers. Our task is to find this distance
from the intersection to the car to the nearest kilometer.

If we view the displacement as a
vector, we can see that the distance we’re looking for is just the size of the
horizontal projection of the displacement. But we can easily find the length
of this vector component by applying the Pythagorean theorem to the right triangle
formed by the northbound road, the eastbound road, and the displacement. This is a right triangle because,
by definition, the direction of north and the direction of east are
perpendicular.

Let’s call the length of this
eastbound road segment 𝑙. Then the Pythagorean theorem tells
us that 10 kilometers squared plus 𝑙 squared, the sum of the squares of the lengths
of the legs of the triangle, is equal to 24 kilometers squared, the square of the
length of the hypotenuse. If we subtract 10 kilometers
squared from both sides, 10 kilometers squared minus 10 kilometers squared on the
left-hand side is zero. And we’re left with 𝑙 squared is
equal to 24 kilometers squared minus 10 kilometers squared. 24 squared is 576, and 10 squared
is 100. So we can rewrite the right-hand
side as 576 kilometers squared minus 100 kilometers squared. 576 minus 100 is 476, so 𝑙 squared
is equal to 476 kilometers squared.

To find 𝑙, we simply take the
square root of both sides. The square root of kilometers
squared is just kilometers. So 𝑙 is equal to the square root
of 476 kilometers. The square root of 476 is
approximately 21.8. Rounding to the nearest kilometer,
we look to the first place to the right of the decimal point, which is eight. Eight is greater than five, so one
rounds up to two. So, to the nearest kilometer, the
distance from the broken-down car to the intersection is 22 kilometers.

Great, let’s see another
example.

A bird flies along a line that
displaces it 450 meters east and 350 meters north of its starting point, as shown in
the diagram. What angle must the bird turn
toward the west to change direction and fly directly north? Give your answer to the nearest
degree.

All right, so we have a diagram
where the cardinal directions are labeled such that east is toward the right and
north is toward the top. The blue line represents the
displacement vector of the bird. And the two black lines represent
the vertical and horizontal components of this displacement. As stated in the question, these
are 350 meters north and 450 meters east, respectively. The question asks us what angle the
bird must turn to change its direction and fly directly towards the north.

To help us figure out this angle,
let’s draw the bird’s northward trajectory on the diagram. This magenta line represents a
northward trajectory since it points directly up. On the other hand, if the bird were
to continue along its current trajectory, it would follow this blue dotted line. So in order to change direction and
fly directly north, the bird has to turn through this angle between the two
trajectories.

This is the angle we need to
calculate. But since we don’t know anything
else about the dotted or magenta vectors, let’s try to find another part of our
diagram with the same angle. The magenta vector and the
northward part of the displacement both point directly north, so they’re
parallel. But this means that the bird’s
trajectory is a straight line intersecting two parallel lines. So corresponding angles at the two
intersections have the same measure.

In our diagram, this angle between
the northern component of the displacement and the displacement itself corresponds
to the angle we drew before, since both of these angles are to the right of one of
the parallel lines and above the displacement line. So if we can find the measure of
this angle, we’ll know the angle that the bird needs to turn.

Now, recall that by adding the
horizontal and vertical components of a vector, we get the vector itself. Therefore, if we draw 450 meters to
the east between the head of the northern vector and the head of the displacement
vector, we’ll get a right triangle formed by the two components and the
displacement. These two vectors form a right
angle because north and east are perpendicular. Let’s call our angle 𝜃.

Now, recall that the tangent of an
angle is equal to the length of the opposite side divided by the length of the
adjacent side. So tan of 𝜃 is equal to 450 meters
divided by 350 meters. Meters divided by meters is just
one. So the right-hand side of this
equation is just a number without units. Now, we can take the inverse
tangent of both sides to find this angle. Plugging into a calculator 𝜃 is
equal to arctan of 450 divided by 350, we get 𝜃 is very close to 52.125
degrees.

Now, remember, this is the same
angle that we were looking for to answer the question. So all we need to do is round 𝜃 to
the nearest degree. Looking to the first digit right of
the decimal point, one is less than five, so two rounds to two. So, to the nearest degree, the bird
needs to turn 52 degrees toward the west to fly directly north.

Okay, now that we’ve seen some
examples, let’s review the key points we’ve learned in this lesson. In this video, we learned that a
full description of the kinematic quantities displacement, velocity, acceleration,
and force must include a magnitude and a direction. This makes vectors the perfect
mathematical object to represent these quantities.

We also saw that a vector
representing any one of these quantities could be represented as a sum of two
vectors, each one parallel to one of the axes in a Cartesian system. Using the Pythagorean theorem and
trigonometry rules for right triangles, we were then able to relate the magnitude
and direction of our original vector to the magnitudes of its components, where
we’ve used the same symbols as the vectors, but without the half arrow on top to
represent the magnitude of each vector.

We were also able to determine the
magnitude of the components in terms of the magnitude and direction of the original
vector. This is sufficient to completely
determine the components because each component’s direction is already known. Using components, we also converted
a vector sum into two scalar sums, one for each component. Using these sums is often quite
easy, which is very useful for us because combining two values for kinematic
quantity can be represented by vector addition.