How to solve this problem involving vectors?

Question

Problem: A particle traveling in a straight line is located at the point $(1,-1,2)$ and has speed $2$ at time $t=0$. The particle moves toward the point $(3,0,3)$ with constant acceleration $2\hat{i}+\hat{j}+\hat{k}$. Find the particle's position vector $\vec{r}(t)$ and the time $t$ that it takes to get to the point $(3,0,3)$.

I did the following:

$$\begin{aligned} &\vec{v}(t)=\int\vec{a}(t)\,dt=\int(2\hat{i}+\hat{j}+\hat{k})\,dt=2t\hat{i}+t\hat{j}+t\hat{k}+\vec{C}\\ &|\vec{v}(0)|=\left|\vec{C}\right|=2\quad\text{assuming}\quad C_1=C_2=C_3\quad\text{then}\quad \vec{C}=\pm\frac2{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\\ &\begin{aligned}\vec{r}(t)&=\int\vec{v}(t)\,dt=\int\left(\left(2t\pm\frac2{\sqrt{3}}\right)\hat{i}+\left(t\pm\frac2{\sqrt{3}}\right)\hat{j}+\left(t\pm\frac2{\sqrt{3}}\right)\hat{k}\right)dt\Big|_{(1,-1,2),\,t=0}\\ &=\left(t^2\pm\frac2{\sqrt{3}}t+1\right)\hat{i}+\left(\frac12t^2\pm\frac2{\sqrt{3}}t-1\right)\hat{j}+\left(\frac12t^2\pm\frac2{\sqrt{3}}t+2\right)\hat{k}\end{aligned}\\ \\ &\begin{cases} t^2\pm\frac2{\sqrt{3}}t+1=3\\ \frac12t^2\pm\frac2{\sqrt{3}}t-1=0\\ \frac12t^2\pm\frac2{\sqrt{3}}t+2=3\\ \end{cases}\implies\text{How can I choose which coefficient for t to take? Is my assumption for these coefficients wrong?} \end{aligned}$$

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Or, the problem says that the particle moves in straight line, so:

$$\begin{aligned}&\vec{u}=(3-1,0-(-1),3-2)=(2,1,1).\\ &p_0=(1,-1,2)\\&\vec{r}(t)=(1+2t)\hat{i}+(-1+t)\hat{j}+(2+t)\hat{k}\\&\vec{v}(t)=\vec{r}(t)'=2\hat{i}+\hat{j}+\hat{k}\\ &\vec{a}(t)=\vec{v}(t)'=0\end{aligned} $$ so this cannot be because conditions are not satisfied. How do I do this? How can the path of a particle be described by a straight line and have acceleration?

Answer 1

We can calculate component by component as follow

• $\vec {v(t)}=2t\hat{i}+t\hat{j}+t\hat{k}+\vec {v_0}$

and from the initial condition

• $\vec {v(t)}=\vec {v_0}=c(2,1,1)$ such that $|\vec {v_0}|=2\implies c=\frac{\sqrt 6}{3}$

then

• $\vec {s(t)}=t^2\hat{i}+\frac{t^2}2\hat{j}+\frac{t^2}2\hat{k}+\vec {v_0} \cdot t +\vec {s_0}$

with $\vec {s_0}=(1,-1,2)$.

The condition to reach $(3,0,3)$ is given by

• $t^2+\frac{\sqrt 6}{3}t+1=3\implies t^2+\frac{\sqrt 6}{3}t-2=0 \implies t=\sqrt{\frac23}$

Answer 2

Your assumption that $C_1=C_2=C_3$, which I presume are the coordinates of $\vec C$, is incorrect. The first constant of integration is simply the velocity at $t=0$, just as in the one-dimensional case. You’re given that the particle moves in a straight line from $P = (1,-1,2)$ to $Q = (3,0,3)$, so its velocity will always be some multiple of $Q-P$, which the vector that you’ve come up with is decidedly not. Here, $\|\vec v(0)\| = 2$, therefore $\vec C = 2{Q-P\over\|Q-P\|} = \sqrt{\frac23}(2,-1,1)$. With the correct initial velocity, you can then proceed to integrate $v(t)$ as you did.

It appears that working with vectors directly might be confusing you. You can instead solve this problem one component at a time. Alternatively, solve the equivalent one-dimensional problem (i.e., transform the coordinate system so that the particle is moving along the $x$-axis), which you know how to do: $\|Q-P\|=\sqrt6$ and $\|(2,1,1)\|=\sqrt6$, so you would compute the equations of motion for a particle moving from the origin towards $x=\sqrt6$ with an initial velocity of $2$ and a constant acceleration of $\sqrt6$. Once you’ve solved that problem, simply transform it back, which is a matter of multiplying the one-dimensional equation of motion by the appropriate unit vector and adding the initial position.