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 Quote by 2x2lcallingcq Problem: Find the position function from the given velocity or acceleration function. a(t) = , v(0)=<4,-2,4>, r(0)=<0,4,-2> Solution: To find the answer the integral must be taken... Integral of a(t) = <-(1/3)e^(-3t), (1/2)t^2, -cos(t)>
Put the constants in right away.
v(t) = <-(1/3)e-3t + c1, (1/2)t2 + c2, -cos(t) + c3>
You're given that v(0) = <4, -2, 4>, so you can solve for the three constants. It's not a simple matter of adding them in as you did.

Do the same thing when you find r(t).
 Quote by 2x2lcallingcq Since taking it with respect to t, it becomes velocity (acceleration = x/t/t distance over time 2) Integral of acceleration with respect to t is... (xt^-1, x/t) Since that is just the velocity function, you do have an initial velocity at v(0)...(time at 0) adding these to each part of the integral. The velocity function is this: <-(1/3)e^(-3t)+4, (1/2)t^2-2, -cos(t)+4> Again, the integral must be taken and it becomes the position function x(t) = <(1/9)e^(-3t)+4t, (1/6)t^3-2t, 4t-sin(t)> And add constants again r(0) therefore the answer r(t) (x(t)) = <(1/9)e^(-3t)+4t+0, (1/6)t^3-2t+4, 4t-sin(t)-2> BUT my teacher said the first and the third portions are wrong? I do not understand how.