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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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eseidlBest ResponseYou've already chosen the best response.1
yeah. You are also given it's initial velocity
 one year ago

snaef999Best ResponseYou've already chosen the best response.0
When finding velocity I need to use the initial velocity as the C right?
 one year ago

eseidlBest ResponseYou've already chosen the best response.1
Yes, the constant of integration after the first integration is your initial velocity
 one year ago

snaef999Best 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)
 one year ago

eseidlBest ResponseYou've already chosen the best response.1
hang on...i'll check and see what I get
 one year ago

eseidlBest ResponseYou've already chosen the best response.1
I get s(t)=(t^3+t)i+(t^4t)j+(t^33t)k
 one year ago

snaef999Best 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.
 one year ago
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