Question

How do you find the velocity and position vectors if you are given that the acceleration vector is `a(t)= (-4 cos (-2t))i + (-4sin (-2t))j + (-2t)k` and the initial velocity is `v(0)=i+k` and the initial position vector is `r(0)= i+j+k`?

Answer 1

First, note that `cos(-2t)=cos(2t)` and `sin(-2t)=-sin(2t)` since cosine is an even function and sine is an odd function.

Now integrate to get `v(t)=-2sin(2t)i-2cos(2t)j-t^2k+\vec{c}`, where `\vec{c}` is a constant of integration. Since `v(0)=i+k` and since `sin(0)=0` and `cos(0)=1`, it follows that `\vec{c}=i+2j+k` so that `v(t)=(-2sin(2t)+1)i+(-2cos(2t)+2)j+(1-t^2)k`.

Now integrate again to get `r(t)=(cos(2t)+t)i+(-sin(2t)+2t)j+(t-\frac{1}{3}t^{3})k+\vec{c}`. Since `r(0)=i+j+k`, it follows that `\vec{c}=j+k` so that `r(t)=(cos(2t)+t)i+(-sin(2t)+2t+1)j+(t-\frac{1}{3}t^{3}+1)k`