I'm taking the AP Calculus BC Exam next week and ran into this problem with no idea how to solve it. Unfortunately, the answer key didn't provide explanations, and I'd really, really appreciate it if someone could explain how to solve this problem.
The acceleration of a particle moving along the line is given by $a(4) = t\cos(t^2)$. If at time $t=0$, its velocity is $2m$ and position is $4m$, what is the position of the particle at time $t=7$?
(Calculator question; the correct answer is $18.303m$ .
I got that the equation for the velocity ($v = \sin(t^2)/2 + 2m$) but I can't seem to find its antiderivative and get the equation for the position.
A step-by-step explanation would be wonderful! :] thank you all.
No comments:
Post a Comment