Video Transcript
In this video, our topic is
position, displacement, and distance. These are fundamental qualities for
many physical situations. In this video, we’ll learn how to
define each of these terms as well as how they relate to one another.
As we get started, say that we have
a bee that’s hovering in place at a certain position. We can really only say what that
position is if we have a set of coordinate axes defined somewhere. Using this as a reference, we can
draw a vector from our origin to the bee’s current position. And further, we can call this
vector 𝐫, specifying where the bee is located relative to this frame.
At this point, let’s imagine that
our bee starts to move, following some path through three-dimensional space. At some later time, if our bee is
positioned here, we can draw a new vector defining that quantity. If we consider the bee’s overall
path, we see that its position vector depends on time. This leads us to rewrite our
position vectors.
Our initial position vector we’ll
say is 𝐫 at a time we’ll call 𝑡 sub i for the initial time. And the final position vector is
here, the position of the bee at some time we’ve called 𝑡 sub f. At any point along its journey, we
could define the bee’s position by drawing a vector from the origin of our
coordinate axes to that particular point on its path. This is the meaning of the bee’s
position.
When the bee’s position changes, as
we see it does all throughout this path, that relates to the term displacement. In this video, we’ll represent
displacement using a lowercase 𝐬. And note that this quantity is a
vector. Displacement, like position, is
typically a function of time. And it’s defined as the change in
an object’s position over time. For example, if we want to
calculate the displacement of our bee at the time we’ve called 𝑡 sub f, then we
would take the bee’s position at that time and subtract from it the bee’s position
at the time we’ve called 𝑡 sub i.
Note that this involves finding the
difference between two vectors. If we sketch that difference in, it
would look like this. Graphically then, we can see the
displacement of the bee over the time interval from 𝑡 sub i to 𝑡 sub f. And notice something interesting
about displacement. It only depends on the starting and
ending position of our object. It doesn’t take into account any
information we have about the path our object followed in traveling between these
two positions.
A quantity that does do this though
is the one we call distance. We’ll represent distance using a
lowercase 𝑑. Like displacement, this is
typically a function of time. But unlike displacement, distance
is not a vector. If we wanted to determine the
distance traveled by our bee over this time interval, we would start at its initial
position. And then we would follow the path
that it followed all throughout every twist and turn until we reached its end
point. And this entire path length is what
we call its distance traveled.
Note then that distance is path
dependent, but displacement is not. For example, if our bee had
followed, say, this path to get from the start to the end point, its displacement
will be the same, but its distance traveled will be different. Another thing to note about these
two terms is that distance is always at least as great as the magnitude of
displacement, if not greater. This is because the shortest
distance between a start and an end point follows along the line that’s defined by
the relevant displacement vector.
Thanks to the fact that
displacement is a vector while distance is a scalar quantity, there’s yet another
difference between these two terms. Because an object’s displacement is
given by a vector, it’s possible for displacement to be negative. For example, say that we have some
particle right here and that over time that particle moves purely in the negative
𝑥-direction according to our coordinate axes. In that case, the particle’s
displacement is defined by this vector, which is negative.
Distance, on the other hand, is
never negative. Any distance traveled is recorded
as a positive value. And so we would say the distance
that this particle has traveled is positive. The least distance that an object
can travel then is zero, if it doesn’t move at all.
Knowing all this about position,
displacement, and distance, let’s get some practice now through a few examples.
Using the given figure, calculate
the distance 𝑑 and the displacement 𝑠 of a body that moves from point 𝐴 to point
𝐶 then returns to point 𝐵.
Looking at this figure, we’re told
to imagine a body that starts here at point 𝐴, moves all the way from point 𝐴 to
point 𝐶, and then doubles back, ending up at point 𝐵. Knowing that the distance this
object travels is represented by 𝑑 while its displacement is represented by 𝑠, we
can recall that the distance an object travels includes the total path length that
it follows. That means that this body’s
distance will include the distance from 𝐴 to 𝐵 to 𝐶 and then back to 𝐵. We see that that’s equal to 28
centimeters plus 24 centimeters then plus 24 centimeters again. That comes out to 76 centimeters in
total.
The displacement of our body, on
the other hand, only takes into account its starting and ending positions. We know that our body starts at
point 𝐴 and it ends up at point 𝐵. That means that its displacement
includes only this length of 28 centimeters. Note that if our body had traveled
just from point 𝐴 directly to point 𝐵, then its distance and displacement would’ve
been the same. We see through the actual scenario
though how these two terms are different.
Let’s now look at an example where
our object of interest moves in two dimensions.
According to the figure, a body
moved from 𝐴 to 𝐵 along the line segment 𝐴𝐵, and then it moved to 𝐶 along
𝐵𝐶. Finally, it moved to 𝐷 along 𝐶𝐷
and stopped there. Find the distance covered by the
body 𝑑 one and the magnitude of its displacement 𝑑 two.
Looking at the figure, we’re told
that a body begins at point 𝐴, right here, and then follows the line segment 𝐴𝐵
to point 𝐵 then moves to point 𝐶 following this path, finally moving along line
segment 𝐶𝐷 to end up at point 𝐷. Given this motion, we want to
calculate the distance the body has covered, we call that 𝑑 one, and the magnitude
of its displacement 𝑑 two.
Clearing some space, let’s first
work on solving for the distance our body travels 𝑑 one. We can recall that distance in
general is equal to the total path length followed by some body as it moves from one
location to another. In our case, our body moved from
point 𝐴 to point 𝐷 along the path shown in orange. We see that that involves 6.6
centimeters of travel from point 𝐴 to point 𝐵, 8.8 centimeters from 𝐵 to 𝐸, 16.4
centimeters from 𝐸 to 𝐶, and on the last leg of the journey 12.3 centimeters. 𝑑 one then equals the sum of these
four distances. Adding them all up gives a result
of 44.1 centimeters. This is the distance our body
traveled.
And now let’s consider its
displacement magnitude 𝑑 two. Displacement is different from
distance in that displacement only takes into account the start point and end point
for some body. In our situation, our body begins
at point 𝐴 and it ends up at point 𝐷. So this straight pink line
connecting these two points represents the displacement magnitude 𝑑 two. We see that the length of this line
segment can be divided up into two parts: one part right here and then the second
part here. Each of these is a hypotenuse of a
right triangle. This means we can use the
Pythagorean theorem to solve for these lengths.
If we call the length of the first
hypotenuse 𝑙 one and that of the second 𝑙 two, then we can say that the
displacement magnitude 𝑑 two is equal to their sum and that 𝑙 one and 𝑙 two are
defined this way. 𝑙 one equals the square root of
6.6 centimeters quantity squared plus 8.8 centimeters quantity squared. And 𝑙 two equals the square root
of 16.4 centimeters quantity squared plus 12.3 centimeters quantity squared. When we enter these expressions on
our calculator, we find that 𝑙 one is equal to exactly 11 centimeters, while 𝑙 two
is 20.5 centimeters. Adding these together gives us a
result of 31.5 centimeters. This is the displacement magnitude
of our body.
Now let’s look at calculating
displacement based on particle position as a function of time.
A particle started moving in a
straight line. After 𝑡 seconds, its position
relative to a fixed point is given by 𝑟 equals quantity 𝑡 squared minus four 𝑡
plus seven meters when 𝑡 is greater than or equal to zero. Find the displacement of the
particle during the first five seconds.
Okay, so here we have a particle’s
position, given as a function of time when 𝑡 is greater than or equal to zero. And based on this, we want to solve
for the particle’s displacement during the first five seconds, that is, from 𝑡
equals zero up to 𝑡 equals five. To do this, we can recall that the
displacement of a particle — we can represent that displacement with a lowercase 𝑠
— at some time we’ll call 𝑡 sub f, is given by the change in the particle’s
position from some initial time 𝑡 sub i to this final time 𝑡 sub f.
This definition tells us that to
calculate the displacement of our particle over the first five seconds, we’ll need
to know its position at 𝑡 equals zero and 𝑡 equals five seconds. We can write that like this. Displacement at 𝑡 equals five
seconds equals position at 𝑡 equals five seconds minus position at 𝑡 equals zero
seconds.
Our given equation for particle
position is valid for both of these times. So we’ll start by solving for the
particle’s position at 𝑡 equals five. This is equal to five squared minus
four times five plus seven meters. And we subtract from this the
particle’s position at 𝑡 equals zero. That’s equal to zero squared minus
four times zero plus seven meters. We know that zero squared is zero,
and so is negative four times zero. So our particle’s position at time
𝑡 equals zero simplifies to seven meters. At five seconds, its position is
five squared, which is 25, minus four times five, which is 20, added to seven
meters. This adds up to 12 meters. So 12 meters minus seven meters
gives us the displacement we want to solve for. It’s five meters. This is the particle’s displacement
during the first five seconds.
Let’s look now at calculating
displacement from a two-dimensional position vector.
If the position vector of a
particle is expressed as 𝐫 equals two 𝑡 plus three 𝐢 plus five 𝑡 minus two 𝐣,
then the magnitude of the displacement at 𝑡 = 2 seconds equals blank length
units.
Here then we want to fill in this
blank by solving for the magnitude of this particle’s displacement at 𝑡 equals two
seconds. We can begin to do this by
recalling that the displacement vector of an object at a time we’ll call 𝑡 sub f is
equal to that object’s position vector at that same time minus the object’s position
vector at some initial time we’ll call 𝑡 sub i.
In our case, as we want to solve
for displacement magnitude at a time of 𝑡 equals two seconds, we can start by
figuring out the displacement vector at this time. According to our definition, this
displacement vector is equal to the position vector at that same time, two seconds,
minus the position vector at a time 𝑡 equals zero. Knowing this, we can then use our
position vector, which we see is given to us as a function of time, to let us solve
for particle displacement at two seconds.
If we plug in 𝑡 equals two to our
equation for the position vector, we get two times two plus three 𝐢, that’s seven
𝐢, plus five times two minus two 𝐣. That’s equal to eight 𝐣. We can then replace this vector
with the expression we’ve just solved for. Next, we want to solve for the
particle’s position at time 𝑡 equals zero seconds. That’s equal to two times zero plus
three 𝐢, that simplifies to three 𝐢, plus five times zero minus two 𝐣. The vector overall then is three 𝐢
minus two 𝐣. And we substitute this in for 𝐫 of
zero.
We’re now ready to calculate the
displacement at a time 𝑡 equals two seconds. And we get a result of four 𝐢 plus
10𝐣. This, however, is not our final
answer because we want to solve for the magnitude of this displacement. Let’s recall that if we have a
two-dimensional vector, we’ll call it 𝐕, then the magnitude of that vector is equal
to the square root of the sum of the squares of its 𝑥- and 𝑦-components. This means that the magnitude of
our particle’s displacement at 𝑡 equals two seconds equals the square root of four
squared plus 10 squared. Four squared is 16, and 10 squared
is 100. So we’re taking the square root of
116.
116 though is divisible by
four. It’s equal to four times 29. And then, knowing that four is
equal to two squared, we can move it outside the square root sign, which makes our
final answer two times the square root of 29. This is the magnitude of our
particle’s displacement at 𝑡 equals two seconds.
Let’s finish up now by reviewing
some key points about position, displacement, and distance. In this video, we saw that a
particle’s position is a vector quantity that depends on the coordinate system being
used to define that position. We then saw that displacement is
also a vector quantity, and it’s equal to a difference in position. And lastly, distance is a scalar
quantity that’s equal to the length of an object’s path, and distance cannot be
negative.
