Video Transcript
In this video, we’re going to talk
about acceleration vectors. What they are, what information
they show us, and how to calculate them knowing velocity or displacement.
To start off, imagine you were
piloting a spaceship through outer space. The goal of your mission is to land
on and explore the planet Galacton five so far unvisited by humankind. But there’s a problem. Because of unexpected resistance as
your spaceship left Earth’s atmosphere, it’s a little bit behind schedule. The original plan was for your
spaceship to intercept Galactron five at a given point on its circular orbit. But now, the spaceship will need to
speed up unexpectedly in order to make that rendezvous point.
Considering that the velocity of
Galactron five, the velocity of the spaceship, and the distance between the ship’s
current location and the intended meeting point is known, what is the acceleration
𝑎 the ship will need to have in order to meet Galactron five at the intended
location? To answer this question, we’ll need
to know more about acceleration vectors.
In general, an object accelerates
whenever it changes velocity. The change in velocity can involve
speeding up while staying in the same direction. This is what we see with 𝑣 one
going to 𝑣 two going to 𝑣 three. In this case, we’re speeding up to
the right. So, our acceleration vector points
that way. On the other hand, we could have a
constant speed, say, as an object moves in uniform circular motion, but the
direction is constantly changing. This also leads to a change in
velocity, which tells us that acceleration is happening.
An acceleration vector shows how
much velocity is changing by how long it is. Considering these two vectors 𝑎
one and 𝑎 two, 𝑎 two because it’s longer represents a greater change in velocity
than 𝑎 one. And an acceleration vector shows
direction by which way it points. Each of these three vectors 𝑎 one,
𝑎 two, and 𝑎 three all have the same magnitude or length, but they’re different
vectors because they point in different directions.
We’ve said that an object
accelerates whenever it changes velocity. We can write this mathematically as
acceleration 𝑎 is equal to Δ𝑣, change in velocity, over the time Δ𝑡 during which
that velocity change happens. Based on this relationship, we can
express acceleration in two different ways. In the first way, we can solve for
an average acceleration. That is, an acceleration that
happens over some time interval. We’ll call it 𝑡 sub 𝑓 minus 𝑡
sub 𝑖. The velocities that correspond to
those two points in time we call 𝑣 sub 𝑓 and 𝑣 sub 𝑖. The difference between those over
the time interval is the average acceleration an object experiences.
On the other hand, if we were to
shrink this time interval 𝑡 sub 𝑓 minus 𝑡 sub 𝑖 smaller and smaller and smaller,
eventually it would become infinitesimally small and we would arrive at
instantaneous acceleration. This is equal to 𝑑𝑣 𝑑𝑡, the
time derivative of velocity. Instantaneous acceleration tells us
an object’s acceleration at a specific point in time.
Looking at this relationship for
instantaneous acceleration, we can recall that it looks similar to our relationship
for instantaneous velocity. The velocity of an object at one
instant in time is equal to the time derivative of its displacement 𝑥. For these three quantities,
acceleration, velocity, and displacement, what connects them is derivatives with
respect to time.
We can say that, given displacement
with respect to time, if we want to solve for velocity or given velocity and we want
to solve for acceleration, then it’s the derivative with respect to time that lets
us make that transition. And going in the other direction,
if we have acceleration and wanna solve for velocity or velocity and wanna solve for
displacement, it’s an integral with respect to time that helps us make that
step. Now that we’ve learnt a bit about
average and instantaneous acceleration, let’s get some practice with these concepts
through a few examples.
A particle accelerates
uniformly. At the time 𝑡 equals 0.0 seconds,
the particle has a velocity 𝑣 equals 14𝑖 plus 22𝑗 meters per second. At 𝑡 equals 3.8 seconds, the
particle has a velocity 𝑣 equals 0.0𝑖 plus 11𝑗 meters per second. What is the acceleration of the
particle?
Since we’re solving for the
acceleration of the particle over a time interval from 𝑡 equals 0.0 to 3.8 seconds,
we know that we’re solving for an average acceleration. To begin on our solution, we can
recall the mathematical relationship for average acceleration. The average acceleration of an
object is equal to its final velocity minus its initial velocity divided by the time
interval over which that velocity change happens.
In our case, we could write that
our initial time 𝑡 sub 𝑖 is 0.0 seconds. Our initial velocity is 14𝑖 plus
22𝑗 meters per second. And our final time is 3.8 seconds,
and our final velocity is 0.0𝑖 plus 11𝑗 meters per second. If we calculate the difference 𝑣
sub 𝑓 minus 𝑣 sub 𝑖, we find, as we treat these vectors separately by their 𝑖
and 𝑗 components, that we end up with a total vector of negative 14𝑖 minus 11𝑗
meters per second.
So, when we go to calculate our
average acceleration, we have this vector as our numerator divided by our time
difference of 𝑡 sub 𝑓, 3.8 seconds, minus 𝑡 sub 𝑖, 0.0 seconds. Our total time interval then is 3.8
seconds. And when we calculate this
fraction, we find a result of negative 3.7𝑖 minus 2.9𝑗 meters per second
squared. That’s the acceleration experienced
by the particle over this time interval.
Now, let’s look at an example that
involves calculating instantaneous acceleration.
A particle has a velocity given by
𝑣 as a function of 𝑡 equals 5.0𝑡𝑖 plus 𝑡 squared 𝑗 minus 2.0𝑡 cubed 𝑘 meters
per second. What is the particle’s acceleration
vector at 𝑡 equals 2.0 seconds? What is the magnitude of the
particle’s acceleration at 𝑡 equals 2.0 seconds?
Since we’re asked to solve for
particle acceleration at a particular time 2.0 seconds, we know that this is an
instantaneous acceleration. We can write down that given time
value 2.0 seconds as well as the particle’s velocity function 𝑣 of 𝑡. We’ll use this information in part
one to solve for instantaneous acceleration and in part two to solve for the
magnitude of that instantaneous acceleration.
We can begin solving for the
instantaneous acceleration by recalling the mathematical equation that explains that
term. The instantaneous acceleration an
object undergoes is equal to its change in velocity divided by its change in time,
specifically the time derivative of its velocity as a function of time. We can write that our acceleration
as a function of time is equal to the time derivative of velocity.
When we plug in for our velocity
equation and take this time derivative, we find it’s equal to 5.0𝑖 plus 2𝑡𝑗 minus
6.0𝑡 squared 𝑘 meters per second squared. This is our generalized solution
for acceleration. But we want to solve for
acceleration at a particular instant in time, when 𝑡 equals 2.0 seconds. To solve for it, we’ll insert that
time value everywhere that 𝑡 appears in our general acceleration equation. When we calculate this value, we
find it’s equal to 5.0𝑖 plus 4.0𝑗 minus 24𝑘 meters per second squared. That’s the acceleration of our
object when 𝑡 equals 2.0 seconds.
Now that we know the particle’s
acceleration at that time, we wanna solve for the magnitude of that
acceleration. That magnitude, which tells us how
much the particle’s velocity is changing at the instant in time 2.0 seconds, is
equal to the square root of the acceleration’s 𝑥-component squared plus its
𝑦-component squared plus its 𝑧-component squared.
When we look at our instantaneous
acceleration expression for these components, we find the 𝑥-component is 5.0, the
𝑦-component is 4.0, and the 𝑧-component is negative 24. When we enter this expression on
our calculator, we find that, to three significant figures, it’s 24.8 meters per
second squared. That’s the magnitude of the
particle’s acceleration when 𝑡 equals 2.0 seconds.
Now, let’s take a moment to
summarize what we’ve learned about acceleration vectors. We’ve seen that an object
accelerates whenever it changes velocity. That is, whenever its speed or its
direction changes. We’ve also seen that, based on the
mathematical definition of acceleration, we can write expressions for average as
well as instantaneous acceleration.
Average acceleration is equal to a
final velocity minus an initial velocity divided by the time interval over which
that change occurs. And instantaneous acceleration is
equal to the time derivative of velocity. And we’ve seen that derivatives and
integrals with respect to time let us move from displacement to velocity to
acceleration and back.
