### Video Transcript

In this video, we’re going to learn
about instantaneous speed and velocity. What these two terms mean, why
they’re instantaneous, and how to use them in solving problems. To introduce this topic, let’s
imagine for a moment that you design roller coasters for amusement parks. And you’re working on a new design
for a roller coaster to be installed this coming year. Designing roller coasters is a
highly technical task. And with this project, you’re
working under a particular constraint. That the maximum velocity the
people on the coasters are allowed to experience at any one point is 100 kilometers
per hour. With this as the upper velocity
limit, you want to find out if there’s any point along the roller coaster’s path
where the velocity exceeds this value. And if so, the design will have to
be modified. How can you figure out if, under
the current design, the coaster’s velocity ever exceeds this maximum upper
limit?

To find out, let’s look into these
ideas of instantaneous velocity and instantaneous speed. As a bit of background, let’s look
at a position versus time plot. On this plot, we have position in
meters on our vertical axis and time in seconds on our horizontal axis. And we have the position over time
of some object plotted out on this chart. Let’s say that we pick a time
somewhere around the halfway mark of the object’s journey. And we say that we’d like to know
the velocity of the object at that time. There are a few ways we could
approach solving for that particular velocity. One way is to consider that point
as the midpoint, roughly, between the start and the end of our object’s journey. We could write that our velocity at
the midpoint of our journey is approximately equal to our object’s position at the
final time minus the object’s position at the initial time of the journey divided by
the final time minus the initial time.

You may recognize this equation as
an expression of average velocity. But if we calculate that average,
we can see that it might be fairly different from the actual instantaneous velocity,
𝑣, at the point we’re interested in. Seeing that difference, we might
think, “well, let’s move our time points of final and initial closer in to our
actual time we’re interested in.” Let’s say we move them in closer so
that now the slope between them, that is, the average velocity between those two
points, more closely matches the instantaneous velocity at the point we want to find
it. But as we look at this setup, we
see we could still do better. We could move our time points in
even closer.

Now we’d move our 𝑡 initial and 𝑡
final until they’re right up against the point we’re interested in. We’re still calculating an average
velocity. But that average is approaching
more and more our instantaneous velocity. This approximation is best of all
so far. Yet, it’s still not exact. Here is what we’ve learned so
far. The average velocity of our object
is equal to its change in position over the change in time. But if we want to compute an
instantaneous velocity, that is, a velocity at a particular time value. That change in time has to approach
zero. It’s only as our initial and final
times approach one another and indeed become infinitely close. That we’re truly calculating an
instantaneous velocity. We can use a special mathematical
notation to express this idea. We can say that an object’s
instantaneous velocity, that is, its velocity at a particular time value, is equal
to the change in its position divided by the change in its time. As that change in time, Δ𝑡, gets
smaller and smaller approaching zero.

There’s another way to express this
idea using what’s called differential notation. We could say that instantaneous
velocity is equal to 𝑑𝑥, where 𝑥 is position, divided by 𝑑𝑡. Built into this notation is the
idea that 𝑑𝑡 is infinitesimally small in this expression. Consider for a moment the
similarities between average velocity and instantaneous. If we take the form of our average
velocity equation and we take our time values and shrink that gap smaller and
smaller and smaller until the gap approaches zero. Then we arrive at our equation for
instantaneous velocity. Practically, when we calculate
instantaneous velocity in example questions, we’ll be taking derivatives of position
as a function of time with respect to that variable time.

We’ve talked so far only about
instantaneous velocity. But what about instantaneous
speed? How are the two similar, and how
are they different? It turns out that instantaneous
speed is equal to the magnitude of instantaneous velocity. Recall that speed is a scalar while
velocity is a vector. So while our instantaneous velocity
could be negative, our instantaneous speed will always be positive or zero. Now that we have an idea for
instantaneous velocity and instantaneous speed, let’s get some practice using these
new concepts.

The position of an object changes
as a function of time according to 𝑥 of 𝑡 equals negative three 𝑡 squared
meters. What is the object’s velocity when
𝑡 equals one second? What is the object’s speed when 𝑡
equals one second?

Given a function describing an
object’s position as a function of time, we want to know its velocity when 𝑡 is
equal to one second and its speed at that same time. In other words, we want to solve
for its instantaneous velocity and instantaneous speed. We can recall that an object’s
instantaneous velocity is equal to the derivative of its position with respect to
time. In our case, we’re given 𝑥 or
position as a function of time and can plug in for that expression. When we take its derivative with
respect to time, we find a result of negative six times 𝑡, now with units of meters
per second. This is the instantaneous velocity
of our object at a general time 𝑡. But we want to solve for that
velocity when time is equal to one second. When 𝑡 is equal to one second, our
instantaneous velocity is negative six times one meters per second, or negative six
meters per second. That’s our instantaneous velocity
when 𝑡 equals one second. Now what about its speed at that
time?

We can recall that instantaneous
speed is a scalar quantity. And it’s equal to the magnitude of
instantaneous velocity. This means that the instantaneous
speed of our object, when 𝑡 is equal to one second, is equal to the absolute value
of negative six meters per second. This simplifies to six meters per
second. That’s the object’s instantaneous
speed when time equals one second.

That’s a bit about the difference
between instantaneous speed and velocity. Now let’s look at an exercise that
highlights the difference between average velocity and instantaneous velocity.

A particle’s position varies
according to 𝑥 as a function of 𝑡 equals 3.0𝑡 squared plus 0.50𝑡 cubed
meters. What is the particle’s
instantaneous velocity when 𝑡 equals 2.0 seconds? What is the particle’s average
velocity between 𝑡 equals 1.0 seconds and 𝑡 equals 3.0 seconds?

In part one of this exercise, we
want to solve for instantaneous velocity when 𝑡 equals 2.0 seconds. Then in part two, we’ll solve for
the average velocity of the particle over a time interval centered on that same time
value. This one begins at 1.0 seconds and
ends at 3.0 seconds. In calculating instantaneous
velocity, we can recall that it’s equal to the time derivative of the position as a
function of time of our particle of interest. So in our case, the instantaneous
velocity of the particle as a function of time equals 𝑑 𝑑𝑡, the time derivative
of position as a function of time which we’re given. When we plug in our equation for
position as a function of time and take the time derivative. This derivative results in the
expression 6.0𝑡 plus 1.50𝑡 squared in units of meters per second.

We’ve arrived at a general
expression for instantaneous velocity. But we want to solve for the
velocity at a particular time value, when 𝑡 equals 2.0 seconds. To calculate that value, we plug in
a value of 2.0 seconds for 𝑡 in our expression. When we calculate this value, to
two significant figures, our result is 18 meters per second. That’s the instantaneous velocity
of our particle when 𝑡 equals 2.0 seconds. Next, we move on to solving for the
average velocity of our particle over the time interval 1.0 to 3.0 seconds. This average velocity will be equal
to the position of our particle at 3.0 seconds minus its position at 1.0 seconds
divided by the time interval 3.0 minus 1.0 seconds. Or, in the denominator, 2.0
seconds.

To solve for the position of our
particle at various times, we can use the expression given to us in the problem
statement. When we plug in a value of 3.0
seconds for 𝑡 and calculate the position, we find a result of 40.5 meters. Which we insert for our position
when 𝑡 equals 3.0 seconds. We then do the same thing for 𝑡
equals 1.0 seconds, plugging that value in to our expression and finding a result of
3.50 meters. Which we then insert for our
position when 𝑡 equals 1.0 seconds. We’re now ready to calculate the
average velocity of our particle over the time interval of interest. When we do and round the result to
two significant figures, we find it’s equal to 19 meters per second. So our average velocity and our
instantaneous velocity, even though they’re centered on the same time values, are
not the same.

Let’s summarize what we’ve learned
about instantaneous speed and instantaneous velocity. Instantaneous speed and
instantaneous velocity are rates at a specific point in time. Instantaneous velocity is equal to
the time rate of change of an object’s position, while instantaneous speed is equal
to the magnitude of instantaneous velocity. And we can think of instantaneous
velocity graphically by considering a position-versus-time curve. And consider that as our change in
time shrinks smaller and smaller and smaller, we’re able to get a more finely
resolved velocity. Until finally, as Δ𝑡 approaches
zero in that limit, we have an instantaneous velocity at a particular time
value. Instantaneous speed and
instantaneous velocity are another tool we can use when analyzing object motion.