Question

Assume the acceleration a(t) = (sin t; 10t; t
2 1) and the
initial position c(0) = (10; 0; 0). Find the position curve c(t) if the
initial velocity is v(0) = (0; 0; 2006)

Answer 1

a(t) = (sin t; 10t; t 2 1), not quite sure with the 2 1 numbers right there fore...but i'll solve this as a(t)=(sin(t), 10t, t)
y= c(t)
y'=v(t)
y''=a(t)
so y''-> y' as a(t) -> v(t)
a(t)= (sin(t), 10t , t)
v(t)= (-cos(t) +c1, 5t^2 +c2, t^2/2 + c3)
v(0)= (-cos(0) + c1, 0 + c2, 0 +c2)= (0,0,2006)
c1= 0 +cos(0)= 0 +1= 1
c2=0
c3= 2006
v(t)= (-cos(t) +1, 5t^2, t^2/2)
v(t)->c(t), as y' -> y
c(t)= (-sin(t) + c1, 5t^3/3 +c2, t^3/6 +c3)
c(0)=(-sin(0) +c1, 0 +c2, 0 +c3)=(10,0,0)
c1= 10 + sin(0)= 10+0=10
c2= 0
c3= 0
c(t)=(-sin(t) +10, 5t^3/3, t^3/6) <--- Answer

Just do the anti-derivative 2 times to go back to original distance equation. Don't forget the constant every time you do it, and use the given conditions to find the constant!