Can we know whether it’s a $1D$ or a $2D$ motion just by looking at the position-time relation? How do I know whether it is a $2D$ or a $1D$ motion, just by looking at position-time, or velocity-time, or acceleration-time equations? 
Maybe the question is not very clear, I’m not sure I’m getting it across properly, so I’ll try to use some examples to make the question clearer. 
We have a position time equation : $\vec r$ = $6t^2$$\hat i$ + $3t^2$$\hat j$ 
It’s easy to see that it is a $1D$ motion, because its locus equation is a straight line. 
Likewise, $\vec r$ = $5t$$\hat i$ + $2t^2$$\hat j$ is a $2D$ motion, because its trajectory equation is a parabola. 
Other examples of two-dimensional motions are : 
$\vec r$ = $30t$$\hat i$ + ($20$ - $10t^2$)$\hat j$    (projectile motion)
$\vec r$ = $sin2t$$\hat i$ + $cos2t$$\hat j$ (Circular motion) 
How did I know that these were $2$-Dimensional motions? I checked their trajectory equations. 
My question is, is it possible to know just by looking at position-time equations, whether the body is moving in a straight line or changing its direction (i.e $2D$ motion), without checking its equation of trajectory? 
 A: In general the position vectors you are looking at take the form $$\mathbf x=f(t)\hat i+g(t)\hat j$$
Now, let's think about what is first taught when learning about lines. In the x-y plane, a line can be described by
$$y=mx+b$$
Now, you can probably convince yourself that our vector components $\langle i,j\rangle$ can be viewed as coordinates $(x,y)$. Therefore, our motion is along a line if
$$g(t)=mf(t)+b$$ for constants $m$ and $b$. (Unless $f$ is a constant function, then $g$ can be anything and we will still have a line without following this form (analagous to lines of the form $x=c$). Except if $f$ and $g$ are both constant functions, then you are just sitting at a point).
A: My solution:
2D case
given the position vector $\vec {R}$ with the parameter $t$
$$\vec{R}=\left[ \begin {array}{c} x\\y\end {array} \right]=\left[ \begin {array}{c} f \left( t \right) \\ g
 \left( t \right) \end {array} \right] 
$$
Ansatz $y(x)=a\,x+b$ 
,the slope $a$ must be const.!, $\quad$ with $a$
$$a=\frac{\partial y}{\partial x}=\frac{\frac{d g}{dt}}{\frac{df}{dt}}=\text{const}\tag 1\quad \frac{df}{dt}\ne 0 $$ 
3D case
$$\vec{R}= \left[ \begin {array}{c} x\\ y\\
z\end {array} \right] 
=\left[ \begin {array}{c} f \left( t \right) \\ g
 \left( t \right) \\ u \left( t \right) \end {array}
 \right] 
$$
Ansatz $z(x,y)=a\,x+b\,y+c$ 
,the slope $a$ and $b$  must be const.!,$\quad$ with $a$
$$a=\frac{\partial z}{\partial x}=\frac{\frac{d u}{dt}}{\frac{df}{dt}}=\text{const}\tag 2,\quad \frac{df}{dt}\ne 0$$
and
$$b=\frac{\partial z}{\partial y}=\frac{\frac{d u}{dt}}{\frac{d g}{dt}}=\text{const}\tag 3,\quad \frac{dg}{dt}\ne 0$$ 
Example
$$\vec{R}=\left[ \begin {array}{c} 6\,{t}^{2}\\ 3\,{t}^{2}
\end {array} \right] 
$$
$a=\frac{1}{2}=\text{const}$ ,linear function
$$\vec{R}= \left[ \begin {array}{c} \sin \left( 2\,t \right) 
\\ \cos \left( 2\,t \right) \end {array} \right]
$$
$a=\tan(2\,t)$ ,nonlinear function
$$\vec{R}= \left[ \begin {array}{c} {t}^{2}-10\\{t}^{2}+5
\\{t}^{2}\end {array} \right] 
$$
$a=1\quad,b=1$ ,linear function  
