### Video Transcript

In this video, we’re going to learn
about average and instantaneous acceleration. What these terms are, how they
relate to velocity and position, and how to use them practically in solving
problems.

As we start off, imagine you’re at
the racetrack watching your favorite driver compete in a championship race. After several hours of driving, the
cars all come down to their final lap around the racecourse. Unfortunately, your favorite driver
is not in front. And you begin to wonder, if all the
other cars maintain their current speed, then would it even be possible for your
driver to accelerate past them and cross the finish line first? This is a question we’ll need to
understand acceleration to be able to answer.

Acceleration is the time rate of
change of velocity. This means that if we wanted to
solve for an object’s average acceleration, then over some time interval, starting
with 𝑡 initial and ending with 𝑡 final, we would subtract the object’s velocity at
the initial time from the object’s velocity at the final time and divide that by the
difference in time total.

If we were to extend this notion to
calculate an instantaneous acceleration, that is acceleration at a specific point in
time, then we would deliberately shrink the gap between the final and the initial
time, smaller and smaller until that time differential approached zero. Now, we know that instantaneous
acceleration is equal to a time derivative of velocity. And we may recall that
instantaneous velocity is given as a time derivative of position.

So, we now have three terms,
position, velocity, and acceleration, that all are connected to one another. We could almost think of it like a
ladder, where position is on the top rung, velocity is on the next rung, and
acceleration is below that. To get from position to velocity,
we would take a derivative with respect to time. And likewise, to get from velocity
down to acceleration. And if we wanted to move up the
rungs of the ladder, say from acceleration to velocity, then we take an integral of
acceleration with respect to time. And similarly, to get from velocity
to position.

As a side note, these three terms
aren’t the only terms that describe motion. If we were to take the time
derivative of acceleration, we arrive at a lesser-known term known as jerk. If we were riding in a car whose
acceleration was changing, then this term, jerk, might be a good description for how
it would feel. That said, position, velocity, and
acceleration cover most of the motion that we’re interested in. Now, that we know how to calculate
average and instantaneous acceleration, let’s get some practice using these
concepts.

A racehorse passing through a
starting gate accelerates from rest to a velocity of 15.0 meters per second due
west, in a time interval of 1.80 seconds. Find the racehorse’s average
acceleration. Assume that east corresponds to
positive displacement.

We want to solve for the average
acceleration of the racehorse over this 1.80-second time interval. We can call that time interval
Δ𝑡. And we can call the final velocity
of the racehorse, 15.0 meters per second, 𝑣 sub 𝑓. If we call 𝑎 sub avg the average
acceleration of the racehorse, we can write that that’s equal to the racehorse’s
final velocity minus its initial velocity over Δ𝑡, the time it took to make that
change in velocity.

Since the horse was passing through
the starting gate, we can assume that 𝑣 sub 𝑖 is zero. And since we’re given 𝑣 sub 𝑓 and
Δ𝑡 in the problem statement, we’re almost ready to solve for 𝑎 sub avg. The one point we want to keep in
mind is that this is a vector. And therefore, it could have
positive or negative value.

We’re told that our velocity, 𝑣
sub 𝑓, is in the west direction but that motion to the east is motion that we can
consider in the positive direction. So, this means we can insert a
minus sign in front of our final velocity of the racehorse. After all, it’s moving in the
negative direction, by our convention. So, the acceleration on average is
negative 15.0 meters per second divided by 1.80 seconds, or negative 8.33 meters per
second squared. That’s the average acceleration of
the racehorse.

Now, let’s look at an exercise that
lets us solve for instantaneous acceleration, as well as instantaneous velocity.

The position of a particle along
the 𝑥-axis varies with time according to the equation 𝑥 as a function of 𝑡 equals
1.5 minus 3.3𝑡 squared meters. What is the velocity of the
particle at 𝑡 equals 2.7 seconds? What is the velocity of the
particle at 𝑡 equals 4.3 seconds? What is the acceleration of the
particle at 𝑡 equals 2.7 seconds? What is the acceleration of the
particle at 𝑡 equals 4.3 seconds?

So, given an equation describing
the position of a particle as a function of time, we want to solve for instantaneous
velocity and instantaneous acceleration at two different time values, 𝑡 equals 2.7
seconds and 𝑡 equals 4.3 seconds. We can name the values we want to
solve for 𝑣 of 2.7 seconds, 𝑣 of 4.3 seconds, 𝑎 of 2.7 seconds, and 𝑎 of 4.3
seconds.

Given this information, we can
recall that there is a relationship between position, velocity, and
acceleration. Instantaneous velocity is equal to
the time derivative of position. And instantaneous acceleration is
equal to the time derivative of velocity, which is also equal to the second time
derivative of position. Since we’re given the position of
our particle with respect to time, let’s differentiate it twice to solve for
instantaneous velocity and instantaneous acceleration.

Instantaneous velocity as a
function of time is equal to the time derivative of 1.5 minus 3.3𝑡 squared
meters. Calculating this derivative, we
find it’s equal to negative 6.6𝑡 meters per second. This tells us that, to solve for
velocity when 𝑡 equals 2.7 seconds and when 𝑡 equals 4.3 seconds, we only need to
plug in those time values into this general expression. When we do, to two significant
figures, the velocity at 2.7 seconds is negative 18 meters per second and the
velocity at 4.3 seconds is negative 28 meters per second.

Now, we move on to solving for
acceleration at these two particular time values. Recalling that acceleration is
equal to the time derivative of velocity, we can write that acceleration as a
function of time is equal to the time derivative of negative 6.6𝑡 meters per
second. We find that this derivative is a
constant, negative 6.6 meters per second squared. This means that for any time value,
that will be our acceleration. Therefore, the acceleration at 2.7
seconds equals the acceleration at 4.3 seconds, which is equal to negative 6.6
meters per second squared. That’s the acceleration at these
two time values.

Let’s take a minute to summarize
what we’ve learned about average and instantaneous acceleration. Acceleration is equal to the time
rate of change of velocity. That is, it’s a vector. Average acceleration is equal to
the velocity of an object at some final time minus the velocity of that same object
at an initial time all divided by that time interval, 𝑡 final minus 𝑡 initial.

Instantaneous acceleration is equal
to the time derivative of velocity, 𝑑𝑣 𝑑𝑡. These two equations are
connected. As the time interval between 𝑡
final and 𝑡 initial gets smaller and smaller approaching zero, average acceleration
approaches instantaneous acceleration.

And acceleration is connected with
velocity and position through derivatives with respect to time. Given one of these three
quantities, we can solve for the others either by integrating or differentiating
with respect to 𝑡. Typically, when we know an object’s
position, velocity, and acceleration as a function of time, we’re able to understand
its motion completely.