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?

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).

• Note: Part of me is not convinced this covers all cases or is always correct. If someone has a counter-example or an example showing a more general form (i.e. motion along a line not taking this form), please let me know! Commented Feb 9, 2019 at 6:16
• Thanks a lot. It makes sense. So basically I need to view the vector components as coordinates. And you said, "if $f$ is a constant function, then $g$ can be anything". If $f$ is a constant function, doesn't it mean the it's a $1D$ motion and the body is just moving along $y$ ?? Because it has no displacement along $x$, doesn't it mean it's a case of $1D$ motion?
– 4d_
Commented Feb 9, 2019 at 6:19
• @πtimese Yes, that's why I said it's motion along a line. What I mean by "unless" is that you will have a line that isn't of this form. Just like how $x=c$ is a line but is not of the form $y=mx+b$ Commented Feb 9, 2019 at 6:21
• @πtimese Please consider upvoting and marking the answer as the accepted answer if sufficiently answers the question. Commented Feb 9, 2019 at 6:34
• @AaronStevans I did upvote it earlier, and just marked it as the accepted answer. Thanks for helping me
– 4d_
Commented Feb 9, 2019 at 6:47

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