Two times negative nine, negative 18 times t to the first power plus eight, derivative of eight t is just eight if we're taking the derivative with respect to t. I haven't used yet, I'll use this green, so let's see. And so the x component, well I just take the derivative of the x component again, and let me find a color So this is going to be equal to, this is going to be equal to, let me give myself some space. And if we wanna find theĪcceleration function, or the vector valued function that gives us accelerationĪs a function of time, well that's just going to be the derivative of the velocity function with respect to time. Three t-squared, and then the derivative of two is just a zero, so I actually have space to write that So the derivative of t to the third with respect to t is three t-squared. So it's three times negative three, so it's negative nine t-squared and then plus two times four is eight, so plus eight t to the first. With respect to time, we're just gonna use So if we wanna take the derivative of the x component here Take the corresponding derivatives of each of the components. So V of t is just going toīe equal to r prime of t, which is going to be equal to, well you just have to The velocity vector's just going to be the derivative of that. Now the key realization is if you have the position vector, well The vertical direction, or the y component. The horizontal direction, this is the component in ![]() Two, times the unit vector in the vertical direction. The horizontal direction, plus t to the third plus Third plus four t-squared times the unit vector in It might be written like, or sometimes people write this as unit vector notation. And this is one form of notation for a vector, another way of writing this, you might be familiar And so you give me any time greater than or equal to zero, I put it in here and I can give you theĬorresponding x and y components. Three t to the third power plus four t-squared, and the y component is t to the third power plus two. ![]() ![]() Us that the x component of our position is negative What is the particle's acceleration vector at time t equals three? Alright, so our position, let's denote that it's a vector valued function, it's gonna be a function of time. Than or equal to zero its position vector is,Īnd they give us the x component and the y component of our position vectors, and Moves in the xy plane so that at any time t is greater I am answering this based off what I understood from the video, so if anyone finds any discrepancies, feel free to correct me. I hope this answers the question you were asking, and that it's still useful, if at all. ![]() The function essentially just tells us about the motion of a particle in two dimensions with respect to time, which is why we take the second derivative with respect to time. The second derivative here tells us that the acceleration is in the upper left direction of the Cartesian plane at t=3 (as it has a negative x coordinate and a positive y coordinate, and leftward is taken as negative according to convention) with a magnitude of 46 units/s^2 in the negative x-direction and 18 units/s^2 in the positive y direction. Here, 'x' represents the amount moved along the x-axis as a function of t, while the y-coordinate 'y' represents the amount moved along the y-axis as a function of t. I believe that we should be taking the second derivative with respect to t as its x-coordinate here is a function of t.
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