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.