Question

一道英文的微积分题

A particle moves with velocity v(t) = 6e^-2t for 0  t  5. Find the total distance travelled by the
particle. What is the acceleration of the particle at the end of 4 seconds? Assuming that at t = 0 the
particle was at the origin, nd the position of the particle at the end of 2 seconds.
不知道该怎么做，麻烦帮帮忙！谢谢！

Answer 1

V(t)=6e^(-2t) (0<=t<=5)

4秒后的加速度a=V'(t)=-12e^(-2t)=-12e^(-2*4)=-12e^(-8)
V(0)=6时视为在原点，2秒后的位置：
S(t)=∫V(t)dt=∫6e^(-2t)dt=-3e^(-2t)
当t=0时，S=-3， （原点)
2秒后即t=2时，
S=(V0)t+(1/2)at²
=6t+(1/2)[-12e^(-2t)]t²
=12+(1/2)(-12e^-4)*4
=12-24e^(-4)
即离原点距离为 12-24e^(-4)-(-3)=15-24e^(-4)

追问

能不能麻烦你这道题也看看，谢谢你了！http://zhidao.baidu.com/question/1603742955509916227.html?quesup2&oldq=1

追答

另一题请再提问，采纳答案后再回答新题。