yego.me
💡 Stop wasting time. Read Youtube instead of watch. Download Chrome Extension

Planar motion example: acceleration vector | Advanced derivatives | AP Calculus BC | Khan Academy


3m read
·Nov 11, 2024

A particle moves in the XY plane so that at any time ( T ) is greater than or equal to zero, its position vector is given. They provide us the X component and the Y component of our position vectors, and they're both functions of time. What is the particle's acceleration vector at time ( T = 3 )?

All right, so our position, let's denote that it's a vector-valued function. It's going to be a function of time; it is a vector. They already told us that the X component of our position is ( -3T^3 + 4T^2 ) and the Y component is ( T^3 + 2 ). So you give me any time greater than or equal to zero, I put it in here, and I can give you the corresponding X and Y components.

This is one form of notation for a vector. Another way of writing this, you might be familiar with engineering notation, it might be written like:

[
\mathbf{R}(T) = -3T^3 \mathbf{i} + 4T^2 \mathbf{j}
]

or sometimes people write this as unit vector notation:

[
-3T^3 \mathbf{u_x} + 4T^2 \mathbf{u_y}
]

This is just denoting the same thing. This is the X component; this is the Y component. This is a component in the horizontal direction; this is a component in the vertical direction, or the Y component.

Now, the key realization is if you have the position vector, well, the velocity vector is just going to be the derivative of that. So, ( \mathbf{V}(T) ) is just going to be equal to ( \mathbf{R}'(T) ), which is going to be equal to... well, you just have to take the corresponding derivatives of each of the components.

So let's do that. If we want to take the derivative of the X component here with respect to time, we're just going to use the power rule a bunch. So it's ( 3 \times -3 ), so it's ( -9T^2 ) and then plus ( 2 \times 4 = 8 ), so plus ( 8T ).

Then, over here for the Y component, the derivative of ( T^3 ) with respect to ( T ) is ( 3T^2 ), and the derivative of 2 is just zero. So actually, I have space to write that: ( 3T^2 ).

All right, and if we want to find the acceleration function, or the vector-valued function that gives us acceleration as a function of time, well, that's just going to be the derivative of the velocity function with respect to time.

So, this is going to be equal to... let me give myself some space. The X component, well, I just take the derivative of the X component again. Let me find a color I haven't used yet; I'll use this green.

So let's see: ( 2 \times 9 = 18T ) raised to the 1st power plus 8. The derivative of ( 8T ) is just 8 if we're taking the derivative with respect to ( T ). And then here in the orange, the derivative of ( 3T^2 ) using the power rule here over and over again gives us ( 2 \times 3 = 6T ).

So, we've just been able to find the acceleration function by taking the derivative of this position vector-valued function twice. Now, I just have to evaluate it at ( T = 3 ).

So, our acceleration at ( T = 3 ) is equal to: in green, it's going to be ( -9 \times 3^2 + 8 ), and then we're going to have ( 6 \times 3 ).

So what does this simplify to? Well, this is going to be equal to... let's see: ( -9 \times 3^2 = -81 ) and ( -81 + 8 = -73 ). Then for the Y component, we have ( 6 \times 3 = 18 ).

Did I do that arithmetic right? So this is ( -81 + 8 ), which would be ( -73 ), and ( 18 ) stays the same.

Yep, there you have it: the acceleration vector at ( T = 3 ) is:

[
(-73, 18)
]

That is its acceleration. That is its acceleration vector at ( T = 3 ).

More Articles

View All
Warren Buffett: How to Make Your First $100,000 (5 Steps)
If you want to make your first or next one hundred thousand dollars, you need to follow these five simple lessons from Warren Buffett. The majority of content out there about Warren Buffett gets it completely wrong. That content focuses on how Warren Buff…
Help Khan Academy supercharge learning
Hi everyone, Sal Khan here from Khan Academy, which you probably know is a not-for-profit with a mission of providing a free, world-class education for anyone, anywhere. And not-for-profit means no one owns Khan Academy. We are a public charity. You own …
Example finding critical t value
We are asked what is the critical value t star (t asterix) for constructing a 98% confidence interval for the mean from a sample size of n, which is equal to 15 observations. So just as a reminder of what’s going on here, you have some population. There’…
YouTube Is Deleting My Channel
Hey guys, So I apologize that this is not my usual video, but I wanted to bring something to everyone’s attention because I have a feeling I’m not the only one dealing with this, and I want to bring awareness to some issues happening right now. First of…
Shifts in demand for labor | Microeconomics | Khan Academy
We are now going to continue our study of labor markets, and in this video we’re going to focus on the demand curve for labor. So, let’s imagine that we’re talking about a market for people who work in the pant-making industry. So each of these firms righ…
LC natural response derivation 3
In the last video, we took a guess at what the solution was for our differential equation, and we came up with an exponential as our guess. As we did the analysis, we developed a characteristic equation. We ended up with a complex answer for one of the ad…