A particle is moving in R 3 so that its acceleration function is a ( t ) = ⟨ 2 t , 1 , 0 ⟩. Find the position function, r(t) of the particle if it starts at the point ( − 5 , 0 , 2 ) with initial velocity v ( t ) = ⟨ 3 , 1 , − 1 ⟩We have v ( t ) = ⟨ 3 , 1 , − 1 ⟩ so r ( t ) = ∫ v ( t ) d t= ∫ ⟨ 3 , 1 , − 1 ⟩ d t which is ⟨ 3 t , t , − t ⟩From here I don't know what to do.

Question

A particle is moving in             R        3   so that its acceleration function is   a  (  t  )  =  ⟨  2  t  ,  1  ,  0  ⟩. Find the position function, r(t) of the particle if it starts at the point   (  −  5  ,  0  ,  2  ) with initial velocity   v  (  t  )  =  ⟨  3  ,  1  ,  −  1  ⟩We have   v  (  t  )  =  ⟨  3  ,  1  ,  −  1  ⟩ so   r  (  t  )  =  ∫  v  (  t  )  d  t=   ∫  ⟨  3  ,  1  ,  −  1  ⟩  d  t which is   ⟨  3  t  ,  t  ,  −  t  ⟩From here I don&#039;t know what to do.

lelapem · Accepted Answer

Let            a      →        (  t  )  =  ⟨  2  t  ,  1  ,  0  ⟩According to the definition            d      v              d      t        =  aIntegrating with respect to time  v  (  t  )  −  v  (  0  )  =      ∫          0        t    a  (  λ  )  d  λSo we have            v      →        (  t  )  =  ⟨  3  ,  1  ,  −  1  ⟩  +  ⟨      t    2    ,  t  ,  0  ⟩  =  ⟨      t    2    +  3  ,  t  +  1  ,  −  1  ⟩Similarly  r  (  t  )  −  r  (  0  )  =      ∫          0        t    v  (  λ  )  d  λAnd then            r      →        (  t  )  =  ⟨  −  5  ,  0  ,  2  ⟩  +  ⟨      t    3        /    3  +  3  t  ,      t    2        /    2  +  t  ,  −  t  ⟩  =  ⟨      t    3        /    3  +  3  t  −  5  ,      t    2        /    2  +  t  ,  −  t  +  2  ⟩

A particle is moving in RR^3 so that its acceleration function is a(t)=⟨2t,1,0⟩. Find the position function, r(t) of the particle if it starts at the point (−5,0,2) with initial velocity v(t)=⟨3,1,−1⟩.

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