A particle moving in the xy-plane has velocity v = (2ti + (3-t^2)j) m/s, where t is in s. What is the particle's acceleration vector at t = 4s?

Video transcript

Hey, everyone today we're doing the problem about derivatives. So we're being told that any vector in a two dimensional coordinate system can be reduced to its components X and Y at time is equal to eight seconds. What should be the particles acceleration vector if its velocity is V is equal to four T I plus five minus T squared meters per second. So acceleration X and let's write this out, acceleration can be defined as the change in velocity with respect to time, right. So in this case, the acceleration vector can actually be found by taking the derivative of the velocity factor. So we can also say that this is equal to third of velocity were directed at the time the director of the velocity vector etcetera is out. So is equal to T V or T T she can say is D because the velocity is therefore four T I where I is representative of the X values or exposition or X axis, I guess you could say minus five T squared J separating this out. This will simply be the derivative of four T.I plus the derivative of five minus T squared oops five minus T squared J. Don't forget our extra appearances. Which one solving that we get, that acceleration will be equal to for I plus we're sorry, minus two T jam meters per second squared. Now, remember we're being told that time is equal to eight seconds. We're trying to find what the acceleration is at eight seconds. So when T is equal to eight seconds, then we get a is equal to For I -2-8 J meters per second squared, Which is equal to four, I minus 16 J meters per second squared. So that will be the particles acceleration at time is equal to eight seconds. If the velocity is as written above or answer choice B I hope this helps and I look forward to seeing you all in the next one.