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  • 5 years ago

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)

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  1. anonymous
    • 5 years ago
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    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!

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