- #1
lylek
- 1
- 0
Hi everyone, first time poster.
I was just hoping somebody could explain instantaneous acceleration to me a bit better than my prof did. I know that we take dv / dt. But I am just having trouble applying it. Does this mean that I am putting the derivative x's position equation on top? and say we are looking at t = 3s, what does d (3) mean.
Thanks for any help.
I was just hoping somebody could explain instantaneous acceleration to me a bit better than my prof did. I know that we take dv / dt. But I am just having trouble applying it. Does this mean that I am putting the derivative x's position equation on top? and say we are looking at t = 3s, what does d (3) mean.
Thanks for any help.