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anonymous
 3 years ago
A particle starts at the origin and initial verlocity is i  j + 3k. Its acceleration is a(t) = 6t i + 12t^2 j + 6t k. Find its position function.
I need help confirming my answer.
Is the correct way to do this to integrate a(t) to get velocity and then integrate velocity to get position?
anonymous
 3 years ago
A particle starts at the origin and initial verlocity is i  j + 3k. Its acceleration is a(t) = 6t i + 12t^2 j + 6t k. Find its position function. I need help confirming my answer. Is the correct way to do this to integrate a(t) to get velocity and then integrate velocity to get position?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah. You are also given it's initial velocity

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0When finding velocity I need to use the initial velocity as the C right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, the constant of integration after the first integration is your initial velocity

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0My final answer is (t^3 + t)i + (t^4 t)j + (2/3 t^3 + 3t)k + c (c = 0 because we started at the origin)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hang on...i'll check and see what I get

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I get s(t)=(t^3+t)i+(t^4t)j+(t^33t)k

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ah. I see what I did wrong. Thank you for your help! I looked at the wrong numbers for k when I was dong position integral. It works for me now.
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