Understanding Vectors: Unit Vectors vs Kinematic Equations

[Mr Davis 97]

I am a little confused about how exactly vectors are used. For example, if we had the two statements

\vec{r} = 3\hat{i}t + 4\hat{j}t

and

\vec{r} = \frac{1}{2}\vec{a}t^{2} + \vec{v_{o}}t + \vec{x_{o}}

What is the difference between the two equations (assuming they are both true)? I know that one uses unit vectors and one is a kinematic equation, but I just don't understand the difference between the two, as they both seemingly relate the object's position with time. Why use one over the other? If we have unit vectors, why do we have equations like the one below it? If anybody could clear this up, I would appreciate it.

 

[olivermsun]

Okay, well first of all look at the first equation, which is expressed using unit vectors:

\vec{r} = 3\hat{i}t + 4\hat{j}t

All this is telling you is that the position vector \vec{r} = \vec{r}(t) is some linear function of time. Another common way to write the above equation is

\vec{r} = (3, 4) t = \vec{v}t,

which is just motion with the constant velocity $\vec{v} (3, 4) = 3\hat{i} + 4\hat{j}$, where the motion starts from the initial position $\vec{r}(0) = \vec{0}$.

The next equation:

\vec{r} = \frac{1}{2}\vec{a}t^{2} + \vec{v_{o}}t + \vec{x_{o}}

also describes motion, but for the case of a constant acceleration $\vec{a}$. An initial velocity $\vec{v_{o}}$ and an initial position $\vec{v_{o}}$ are also specified.

If the velocity is constant ($\vec{v} = \vec{v_{o}}$ for all time) then that means $\vec{a} =0$, and the equation reduces to

\vec{r}(t) = \vec{v}t + \vec{x_{o}}.

If the starting position $\vec{x_{o}} = \vec{0}$, then it becomes exactly the same as the first equation above. So actually the first equation (constant velocity, starting point $\vec{0}$) is just a special case of the second equation (constant acceleration, some starting point possibly not at the origin).