Video Transcript
In this video, we will learn how to
determine the acceleration of objects that change the speed at which they move. Acceleration has to do with change
in speed. And to learn about this idea, say
we have an object here on a track along which the object can move. So depending on its motion, the
object will appear at different positions along the track at different times. Let’s say that at zero seconds of
time the object is here, and then as time passes, it remains there. So at one second, two seconds,
three seconds and so on, the object stays in place. Based on this, we know the object
is moving at a constant speed, and that speed is zero. So the object just stays in
place. That’s not so interesting.
But now say we have another object
whose motion we’re following. At first, this object too is at the
left end of the track. But then after one second passes,
the object is here. After two seconds, it’s here. After three seconds, its position
is here and so on, on down the track. Over each one second interval, the
distance this object moves is the same. That is, the distance it travels
between zero seconds and one second is the same as the distance it travels between
one second and two seconds and so on for each additional second. This object then is moving at a
constant speed. But that speed is not zero.
But now, what if we had yet another
object whose position over time looked like this. Just like the other objects, it
starts out here at zero seconds. But then these are its positions
along the track at one, two, and three seconds of time. What can we say about this object’s
motion? To look into it, let’s put marks on
our track to indicate distance. Let’s let meters on the track look
like this. So after one second, our object has
traveled one meter; after two seconds, it’s traveled four meters; and after three
seconds, nine meters. Knowing this, we can calculate our
object’s speed over each one second time interval. Over the first second, our object
moves a distance of one meter. So between zero and one second, our
object’s average speed is one meter per second.
But then let’s look between one
second and two seconds of time. Over this span, our object goes
from one meter up to four meters of distance. That’s a difference of three meters
in one second. This proves that the speed of our
object is changing over time. It’s not a constant speed like it
was in our first two examples. Whenever an object experiences a
changing speed, that means that that object is accelerating. In fact, we can define acceleration
as a change in speed over time.
This expresses in words what we can
also write as a mathematical equation. If we use the letter 𝑣 to
represent speed, the letter 𝑡 to represent time, the letter 𝑎 to represent
acceleration, and then if we symbolize a change using the Greek letter Δ, then we
can combine these symbols so that 𝑎 equals Δ𝑣 over Δ𝑡. Mathematically, this means the same
thing as these words. Acceleration is a change in speed
over time.
Knowing this, let’s look again at
those objects from earlier. Recall that this object here wasn’t
moving at all. It had a constant speed of
zero. This means that for this object,
Δ𝑣, the change in speed is zero, and that means so is its acceleration. Now, what about this object? This object did move at a nonzero
speed, but remember that that speed was constant. Even though the object is moving,
its change in speed is still zero. This object also has an
acceleration of zero. For our last object, though, we
know that there is a change in its speed over time. We saw that the object’s average
speed went from one meter per second to three meters per second.
So let’s solve for this last
object’s acceleration. Its change in speed, Δ𝑣, is three
meters per second minus one meter per second. And then, what is Δ𝑡, its change
in time? At first, we might think that it’s
this time here, two seconds, minus this starting time here, zero seconds. But we must be careful. The change in speed from one up to
three meters per second actually took place over this time interval from one second
to two seconds. That’s because it’s at this moment
after one second has passed that we can say that our object’s average speed is one
meter per second. This means that Δ𝑡 in our equation
is two seconds minus one second. Our fraction then equals three
minus one or two meters per second in the numerator and two minus one or one second
in the denominator.
Now, notice the units in this
expression. There are meters per second divided
by seconds or meters per second per second. These units show us that we really
are calculating a change in speed over time. In this case, the units of speed
are meters per second, and the units of time are seconds. There’s a way we can simplify these
units a bit. Let’s multiply the denominator of
this fraction by one divided by seconds. And then, since we’re doing that to
the denominator, we do the same thing to the numerator. In the denominator, when we
multiply seconds by one over seconds, that’s like multiplying a variable 𝑥 by one
divided by 𝑥. These two values multiplied
together equal seconds over seconds. And that’s just equal to one.
Then in the numerator, we multiply
meters by one and seconds by seconds. That equals meters divided by
seconds times seconds or meters divided by seconds squared. Because this one in the denominator
doesn’t change the fraction, we can leave it out. Our units simplify to meters per
second squared. All this means then that this
fraction here equals two divided by one meter per second squared or just two meters
per second squared. This is the acceleration of our
third object over the first two seconds of its motion.
By working this example, we see
there’s another way to write this general equation for acceleration. We can put it in terms of an
initial and a final speed of our object. We did this when we used three
meters per second and one meter per second earlier. Now, just as there’s such a thing
as constant speed, a speed that doesn’t vary over time, so there is constant or
uniform acceleration. Clearing some space, let’s now see
if our third object maintains this acceleration of two meters per second squared
over all of its motion.
We remember that distances were
marked out on our track like this. We’re now studying this third
interval of time from two to three seconds. Over this one second interval, our
object moved five meters, nine minus four. Its average speed, then, is five
meters per second, over this particular one second interval. We can now calculate the
acceleration of our object between a time of two seconds and the time of three
seconds. That acceleration will equal our
final speed of five meters per second minus the initial speed of three meters per
second divided by a time interval of three seconds minus two seconds. Five minus three is two, and three
minus two is one. So once again, we calculate an
acceleration of two meters per second squared.
We can say then that the same
acceleration still applies to our object motion. So our object has a uniform or
constant acceleration. Its speed is changing, but it
changes the same amount over every additional second of time. That is, it went from one to three
meters per second, a difference of two meters per second over one second, and then
from three to five meters per second over one second, once again a difference of two
meters per second. Knowing all this about
acceleration, let’s look now at a few examples.
Fill in the blank. An object is accelerating when its
speed is (A) increasing, (B) decreasing, (C) increasing or decreasing.
Let’s say we had an object here
whose speed over time is increasing. This means if we looked at the
position of the object at one second intervals, the distances between the positions
of the object would increase. We can recall that acceleration is
defined as a change in speed over a change in time. Therefore, this object whose speed
is increasing over time is accelerating. But notice that for an object’s
speed to change, it doesn’t necessarily need to increase.
Say we had a second object here and
that we knew the positions of this object at zero, one, two, and three seconds. In this instance, the distances
between the object positions with time are getting smaller. But still the speed of the object,
Δ𝑣, is changing. We can say that the speed of the
pink object is decreasing. The particular name given to an
object whose speed decreases over time is deceleration. We can think of deceleration as
negative acceleration. It’s a type of acceleration. Since an object’s speed will change
whether it’s speeding up or slowing down, we can answer that an object is
accelerating when its speed is increasing or decreasing.
Let’s look now at another
example.
Which of the following is a correct
unit for acceleration? (A) Meters per second, (B) meters
per second squared, (C) meters per second quantity squared.
To begin figuring this out, let’s
remember what acceleration is. Mathematically, acceleration is a
change in speed, Δ𝑣, divided by a change in time. So let’s say we had an object which
at a time of zero seconds was moving at a speed of four meters per second. And then two seconds later, the
object is moving at seven meters per second. We can calculate the object’s
acceleration using this relationship. We’ll do this so we can see what
the units of acceleration can be. All right, in the numerator of this
fraction, we want to put the change in our object’s speed. That will be equal to its final
speed, seven meters per second, minus its initial speed of four meters per
second.
Next, let’s think about Δ𝑡, the
time that has elapsed. This is equal to two seconds minus
zero seconds. So this fraction, the acceleration
of our object, is equal to three meters per second divided by two seconds. And here, we’re really only paying
attention to the units. We have the units of speed, meters
per second, divided by the units of time, seconds. The question is, to which one of
our three answer options does this correspond? We can see right away that we won’t
choose option (A). A unit of meters per second is a
speed, not a change in speed over time.
To see how to choose between
options (B) and (C), let’s work on this expression a bit. Right now, we have a fraction here
and we have a fraction here. We can make it though so that we’re
only working with one overall fraction. To do that, we’re going to multiply
the top and bottom of this overall fraction by the same value. We can choose this value to be
whatever we want.
But if we make a specific choice
and choose it to be one over seconds, then when we multiply in the denominator, we
get seconds divided by seconds, which simplifies to one. Then in the numerator, we multiply
meters by one and seconds by seconds. That gives meters divided by
seconds times seconds or meters per second squared. Dividing meters per second squared
by one doesn’t change the value at all. So we’re free to remove the
one. This leaves us with meters per
second squared, option (B).
Note that this is different from
option (C). In option (C), we see meters per
second quantity squared, which means we square both numerator and denominator. That gives us meters squared per
second squared, while a correct unit for acceleration is meters per second
squared.
Let’s look now at one last
example.
A car starts accelerating uniformly
from rest. After accelerating for three
seconds, the car has a speed of 18 meters per second. What is the acceleration of the
car?
Let’s say that this dot here is our
car. And if we start counting time at
zero seconds, we know that at that start since the car begins from rest, it has a
speed of zero meters per second. But then, three seconds later,
we’re told that the car has a speed of 18 meters per second. To begin solving for acceleration,
let’s recall the mathematical equation for acceleration. The acceleration 𝑎 of an object
equals its final speed, we’ll call it 𝑣 two, minus its initial speed all divided by
the amount of time it takes for the object to change speeds from 𝑣 one to 𝑣
two.
In our case, the initial speed of
the object 𝑣 one is zero meters per second, 𝑣 two is 18 meters per second, and the
change in time over which this change in speed happens is three seconds minus zero
seconds or simply three seconds. So then, here’s what the equation
for our object’s acceleration looks like: 18 meters per second minus zero meters per
second is 18 meters per second. And 18 divided by three is six. So when we simplify our units, we
get an answer of six meters per second squared. This is the acceleration of the
car.
Let’s finish our lesson now by
reviewing a few key points. In this video, we learned that
acceleration is a change in speed over time. As an equation, 𝑎 equals Δ𝑣 over
Δ𝑡 or 𝑣 two minus 𝑣 one over Δ𝑡. We also learned that acceleration
is measured in units of meters per second squared, that is, a speed in meters per
second divided by a time in seconds. And lastly, we learned that uniform
acceleration means an object’s speed changes by equal amounts over equal time
intervals. This is a summary of
acceleration.
