WEBVTT
00:00:00.600 --> 00:00:04.200
In this video, we’re going to talk about acceleration vectors.
00:00:04.640 --> 00:00:12.680
What they are, what information they show us, and how to calculate them knowing velocity or displacement.
00:00:13.440 --> 00:00:19.440
To start off, imagine you were piloting a spaceship through outer space.
00:00:20.200 --> 00:00:28.680
The goal of your mission is to land on and explore the planet Galacton five so far unvisited by humankind.
00:00:29.080 --> 00:00:30.600
But there’s a problem.
00:00:31.200 --> 00:00:42.840
Because of unexpected resistance as your spaceship left Earth’s atmosphere, it’s a little bit behind schedule.
00:00:43.240 --> 00:00:52.880
The original plan was for your spaceship to intercept Galactron five at a given point on its circular orbit.
00:00:53.400 --> 00:01:03.560
But now, the spaceship will need to speed up unexpectedly in order to make that rendezvous point.
00:01:04.360 --> 00:01:27.680
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?
00:01:28.560 --> 00:01:34.600
To answer this question, we’ll need to know more about acceleration vectors.
00:01:35.320 --> 00:01:41.680
In general, an object accelerates whenever it changes velocity.
00:01:42.400 --> 00:01:49.600
The change in velocity can involve speeding up while staying in the same direction.
00:01:50.240 --> 00:01:56.240
This is what we see with 𝑣 one going to 𝑣 two going to 𝑣 three.
00:01:56.880 --> 00:02:00.400
In this case, we’re speeding up to the right.
00:02:01.040 --> 00:02:05.280
So, our acceleration vector points that way.
00:02:05.960 --> 00:02:15.960
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.
00:02:16.380 --> 00:02:27.040
This also leads to a change in velocity, which tells us that acceleration is happening.
00:02:27.800 --> 00:02:34.880
An acceleration vector shows how much velocity is changing by how long it is.
00:02:35.400 --> 00:02:49.360
Considering these two vectors 𝑎 one and 𝑎 two, 𝑎 two because it’s longer represents a greater change in velocity than 𝑎 one.
00:02:49.880 --> 00:02:55.680
And an acceleration vector shows direction by which way it points.
00:02:56.360 --> 00:03:11.600
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.
00:03:12.440 --> 00:03:19.320
We’ve said that an object accelerates whenever it changes velocity.
00:03:19.920 --> 00:03:33.880
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.
00:03:34.480 --> 00:03:40.840
Based on this relationship, we can express acceleration in two different ways.
00:03:41.680 --> 00:03:46.120
In the first way, we can solve for an average acceleration.
00:03:46.480 --> 00:03:50.840
That is, an acceleration that happens over some time interval.
00:03:51.520 --> 00:03:55.080
We’ll call it 𝑡 sub 𝑓 minus 𝑡 sub 𝑖.
00:03:55.520 --> 00:04:02.040
The velocities that correspond to those two points in time we call 𝑣 sub 𝑓 and 𝑣 sub 𝑖.
00:04:02.040 --> 00:04:11.680
The difference between those over the time interval is the average acceleration an object experiences.
00:04:12.160 --> 00:04:26.880
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.
00:04:27.400 --> 00:04:33.840
This is equal to 𝑑𝑣 𝑑𝑡, the time derivative of velocity.
00:04:34.360 --> 00:04:40.600
Instantaneous acceleration tells us an object’s acceleration at a specific point in time.
00:04:41.760 --> 00:04:51.200
Looking at this relationship for instantaneous acceleration, we can recall that it looks similar to our relationship for instantaneous velocity.
00:04:51.720 --> 00:04:59.720
The velocity of an object at one instant in time is equal to the time derivative of its displacement 𝑥.
00:05:00.520 --> 00:05:12.080
For these three quantities, acceleration, velocity, and displacement, what connects them is derivatives with respect to time.
00:05:12.480 --> 00:05:32.800
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.
00:05:33.360 --> 00:05:50.160
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.
00:05:50.840 --> 00:06:00.840
Now that we’ve learnt a bit about average and instantaneous acceleration, let’s get some practice with these concepts through a few examples.
00:06:01.600 --> 00:06:04.480
A particle accelerates uniformly.
00:06:05.080 --> 00:06:17.280
At the time 𝑡 equals 0.0 seconds, the particle has a velocity 𝑣 equals 14𝑖 plus 22𝑗 meters per second.
00:06:18.200 --> 00:06:31.240
At 𝑡 equals 3.8 seconds, the particle has a velocity 𝑣 equals 0.0𝑖 plus 11𝑗 meters per second.
00:06:32.000 --> 00:06:33.680
What is the acceleration of the particle?
00:06:34.520 --> 00:06:49.000
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.
00:06:49.520 --> 00:06:57.280
To begin on our solution, we can recall the mathematical relationship for average acceleration.
00:06:57.880 --> 00:07:11.040
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.
00:07:11.920 --> 00:07:19.520
In our case, we could write that our initial time 𝑡 sub 𝑖 is 0.0 seconds.
00:07:19.960 --> 00:07:27.560
Our initial velocity is 14𝑖 plus 22𝑗 meters per second.
00:07:28.040 --> 00:07:40.720
And our final time is 3.8 seconds, and our final velocity is 0.0𝑖 plus 11𝑗 meters per second.
00:07:41.280 --> 00:08:03.400
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.
00:08:04.200 --> 00:08:21.560
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.
00:08:22.200 --> 00:08:27.600
Our total time interval then is 3.8 seconds.
00:08:28.200 --> 00:08:41.000
And when we calculate this fraction, we find a result of negative 3.7𝑖 minus 2.9𝑗 meters per second squared.
00:08:41.720 --> 00:08:47.440
That’s the acceleration experienced by the particle over this time interval.
00:08:48.200 --> 00:08:53.560
Now, let’s look at an example that involves calculating instantaneous acceleration.
00:08:54.280 --> 00:09:07.840
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.
00:09:08.320 --> 00:09:12.520
What is the particle’s acceleration vector at 𝑡 equals 2.0 seconds?
00:09:12.520 --> 00:09:18.120
What is the magnitude of the particle’s acceleration at 𝑡 equals 2.0 seconds?
00:09:19.000 --> 00:09:30.640
Since we’re asked to solve for particle acceleration at a particular time 2.0 seconds, we know that this is an instantaneous acceleration.
00:09:31.200 --> 00:09:40.960
We can write down that given time value 2.0 seconds as well as the particle’s velocity function 𝑣 of 𝑡.
00:09:41.680 --> 00:09:51.640
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.
00:09:52.360 --> 00:10:00.760
We can begin solving for the instantaneous acceleration by recalling the mathematical equation that explains that term.
00:10:01.200 --> 00:10:16.800
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.
00:10:17.680 --> 00:10:23.920
We can write that our acceleration as a function of time is equal to the time derivative of velocity.
00:10:24.720 --> 00:10:40.440
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.
00:10:41.480 --> 00:10:43.720
This is our generalized solution for acceleration.
00:10:44.200 --> 00:10:51.800
But we want to solve for acceleration at a particular instant in time, when 𝑡 equals 2.0 seconds.
00:10:52.480 --> 00:11:01.200
To solve for it, we’ll insert that time value everywhere that 𝑡 appears in our general acceleration equation.
00:11:01.920 --> 00:11:14.120
When we calculate this value, we find it’s equal to 5.0𝑖 plus 4.0𝑗 minus 24𝑘 meters per second squared.
00:11:14.720 --> 00:11:19.120
That’s the acceleration of our object when 𝑡 equals 2.0 seconds.
00:11:20.120 --> 00:11:27.120
Now that we know the particle’s acceleration at that time, we wanna solve for the magnitude of that acceleration.
00:11:27.440 --> 00:11:46.000
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.
00:11:46.600 --> 00:12:02.360
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.
00:12:03.200 --> 00:12:16.800
When we enter this expression on our calculator, we find that, to three significant figures, it’s 24.8 meters per second squared.
00:12:17.400 --> 00:12:22.520
That’s the magnitude of the particle’s acceleration when 𝑡 equals 2.0 seconds.
00:12:23.440 --> 00:12:27.160
Now, let’s take a moment to summarize what we’ve learned about acceleration vectors.
00:12:27.760 --> 00:12:33.360
We’ve seen that an object accelerates whenever it changes velocity.
00:12:33.760 --> 00:12:37.720
That is, whenever its speed or its direction changes.
00:12:38.400 --> 00:12:50.800
We’ve also seen that, based on the mathematical definition of acceleration, we can write expressions for average as well as instantaneous acceleration.
00:12:51.320 --> 00:13:02.920
Average acceleration is equal to a final velocity minus an initial velocity divided by the time interval over which that change occurs.
00:13:03.640 --> 00:13:09.000
And instantaneous acceleration is equal to the time derivative of velocity.
00:13:09.600 --> 00:13:21.160
And we’ve seen that derivatives and integrals with respect to time let us move from displacement to velocity to acceleration and back.