I've been solving initial value problems but this one has me stumped.
So far what I have is that the position when r(0) is $(8,6,5)$ My next given is when $v(0)=8$ Next, the particle moves in a straight line to $(9,8,4)$ I've calculated that the direction vector between $P1$ and $P2$ is the same as the acceleration vector which is my next given. Acceleration is constant $a(t)=i+2j-k$
My thought process has led me so far to 1) Integrate the acceleration vector: $$P_0(8,6,5), P_1(9,8,4)$$ $$P_1P_2=a(t)$$ $$a(t)=i+2j-k$$ $$\int{a(t)dt}=v(t)=(t+C_1)i+(2t+C_2)j-(t+C_3)k$$ $$|a(t)|=\sqrt6$$ $$|v(0)|=8$$ $$v(0)=((0)+C_1))i+(2(0)+C_2)j-((0)+C_3)k$$ $$C_1+C_2-C_3=8$$ This is as far as I've gotten, as I'm not sure how to get to the next step.
Hint: Since the particle moves in a straight line, $v(t)$ must have the same direction vector as $a(t)$, which is $P_1-P_0$. Since you're given $|v(0)|=8$, can you figure out what $v(0)$ is and use that to compute $C_1$, $C_2$, and $C_3$?
Edit: The velocity vector is $s(t)\langle 1, 2, -1\rangle$, where $s(t)$ is the speed, since it is in the same direction as the acceleration vector. Then $$|v(0)| = s(0)\sqrt{6} = 8,$$ so that $s(0) = \frac{4}{3}\sqrt{6}$. This means that $$v(0) = \frac{4}{3}\sqrt{6}\langle 1, 2, -1\rangle.$$ Does that help?