Patricia Bean

2022-07-15

A particle is moving in ${\mathbb{R}}^{3}$ so that its acceleration function is $a\left(t\right)=⟨2t,1,0⟩$. Find the position function, r(t) of the particle if it starts at the point $\left(-5,0,2\right)$ with initial velocity $v\left(t\right)=⟨3,1,-1⟩$
We have $v\left(t\right)=⟨3,1,-1⟩$ so $r\left(t\right)=\int v\left(t\right)dt$= $\int ⟨3,1,-1⟩dt$ which is $⟨3t,t,-t⟩$
From here I don't know what to do.

lelapem

Expert

Let
$\stackrel{\to }{a}\left(t\right)=⟨2t,1,0⟩$
According to the definition
$\frac{dv}{dt}=a$
Integrating with respect to time
$v\left(t\right)-v\left(0\right)={\int }_{0}^{t}a\left(\lambda \right)d\lambda$
So we have
$\stackrel{\to }{v}\left(t\right)=⟨3,1,-1⟩+⟨{t}^{2},t,0⟩=⟨{t}^{2}+3,t+1,-1⟩$
Similarly
$r\left(t\right)-r\left(0\right)={\int }_{0}^{t}v\left(\lambda \right)d\lambda$
And then
$\stackrel{\to }{r}\left(t\right)=⟨-5,0,2⟩+⟨{t}^{3}/3+3t,{t}^{2}/2+t,-t⟩=⟨{t}^{3}/3+3t-5,{t}^{2}/2+t,-t+2⟩$

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