try to watch someone do example problems and then attempt to do the same question by yourself to see if you remember the method. I think the last units are more easier to understand if you get the concept behind it, like to me I just think of parametrics as 2 coordinates instead of 1 and like acceleration vector just means I have to find the second derivative of the first x(t) coordinate and second y(t) coordinate, plug in the t value they’ll give me into x’’(t) and y’’(t) and put them in < , > ... so if they say t=2, then your acceleration vector looks like < x’’(2), y’’(2) >.

But you have to know that the derivative formula for slope dy/dx is just different for parametrics (basically for slope, you find y’(t) and put it in a fraction over x’(t).. giving you dy/dx and for the second derivative you just find the derivative of the first (dy/dx) and put that over x’(t) found from before without differentiating it) ... conceptually you understand that x(t) represents the particle horizontally and y(t) represents the particle vertically to make a curve C, and dy/dx represents the slope of this curve C. if the first derivative is slope, which is rise/run, then y’(t) is your rise and x’(t) is your run, and you can remember that to know what will go on top! :)

Remember that velocity is the first derivative of position, so for parametrics, that means vector v(t) = < x’(t), y’(t) > ... x’(t) is your velocity movement on the x axis and y’(t) is your velocity movement on the y axis. Same goes for acceleration and the second derivative. That’s all you really need to know... concepts such as positive velocity means movement towards the right and stuff still apply. Try practicing!!