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snaef999
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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?
 2 years ago
 2 years ago
snaef999 Group Title
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?
 2 years ago
 2 years ago

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eseidl Group TitleBest ResponseYou've already chosen the best response.1
yeah. You are also given it's initial velocity
 2 years ago

snaef999 Group TitleBest ResponseYou've already chosen the best response.0
When finding velocity I need to use the initial velocity as the C right?
 2 years ago

eseidl Group TitleBest ResponseYou've already chosen the best response.1
Yes, the constant of integration after the first integration is your initial velocity
 2 years ago

snaef999 Group TitleBest ResponseYou've already chosen the best response.0
My 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)
 2 years ago

eseidl Group TitleBest ResponseYou've already chosen the best response.1
hang on...i'll check and see what I get
 2 years ago

eseidl Group TitleBest ResponseYou've already chosen the best response.1
I get s(t)=(t^3+t)i+(t^4t)j+(t^33t)k
 2 years ago

snaef999 Group TitleBest ResponseYou've already chosen the best response.0
Ah. 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.
 2 years ago

eseidl Group TitleBest ResponseYou've already chosen the best response.1
cool, no prob
 2 years ago
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