A particle is moving with acceleration a(t) = cos(t) + sin(t). Given that its position s
satises the initial conditions s(0) = 0 and s(pi/2)=2, find its position at time t =pi .
I know a(t) = f ''(t) =cos(t)+sin(t)
v(t) = f '(t) = sin(t) - cos(t) + C
s(t) = f (t) = -cos(t) - sin(t) +C +D
But what is C, and how to calculate it??
急!微積分問題 Acceleration
2013-06-03 1:02 pm
回答 (2)
2013-06-03 4:00 pm
✔ 最佳答案
since s(0) = 0 and s(pi/2) = 2,you can find C and D by substituting
t = 0, s(t) = 0; t = pi/2, s(t) = 2 into s(t) = -cos(t) - sin(t) + Ct + D
careful not "C", should be "Ct" !!!!!!
參考: knowledge
2013-06-04 2:07 am
Q2 :
Let g(x) = 6 - f ' (x) . Then g(x) >= 0 for all x .
Integration on both sides yields
int(0 to 3) g(x) dx >= 0
int(0to3) ( 6 - f ' (x) ) dx >= 0
6*3 - [ f(3) - f(0) ] >= 0
18 - f(3) + 1 >= 0
f(3) <= 19
Let g(x) = 6 - f ' (x) . Then g(x) >= 0 for all x .
Integration on both sides yields
int(0 to 3) g(x) dx >= 0
int(0to3) ( 6 - f ' (x) ) dx >= 0
6*3 - [ f(3) - f(0) ] >= 0
18 - f(3) + 1 >= 0
f(3) <= 19
收錄日期: 2021-04-13 19:29:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130603000051KK00054