微积分 应用题

Question

一个速度为V的物体在直线上运动 3v=t^2-10t+36 ，找到两点的距离 当t=0和 t=2.

Answer 1

3v=t^2-10t+36,v=1/3*t^2-10/3*t+12.
积分,s=1/9*t^3-5/3*t^2+12t+C.
把t=0和t=2代入上式,相减即可.

Answer 2

dS＝Vdt=(t^2-10t+36)dt/3

积分得：S=(t^3/3-5t^2+36t)/3

S(2)-S(0)=S(2)=(8/3-20+72)/3=164/9

Answer 3

由ds/dt=v可得:
s(t)=vdt=((t^3)/3-5(t^2)+36t)/3
s(2)-s(0)=164/9本回答被提问者采纳

Answer 4

v=(t^2)/3-(10t)/3+36/3
derivative
dv/dt=2/3t-10/3
t=0, v=-10/3
t=2, v=4/3-10/3=-2

-2-(-10/3)=4/3

Answer 5

3v=t^2-10t+36,v=1/3*t^2-10/3*t+12.
积分,s=1/9*t^3-5/3*t^2+12t+C.