Generalization of straight line motion under constant acceleration

Question

My question is that, we all know the three equations of straight line motion under constant acceleration,

\begin{align} x & =x_{\rm o}+v_{\rm o}\,t+\tfrac12 \mathrm a\,t^2 \tag{1d-a}\label{1d-a}\\ v & =v_{\rm o}+\mathrm a\,t \tag{1d-b}\label{1d-b}\\ v^2 & =v_{\rm o}^2+2\,\mathrm a\left(x-x_{\rm o}\right) \tag{1d-c}\label{1d-c} \end{align}

Is my generalization correct? \begin{align} \mathbf r & =\mathbf r_{\rm o}+\boldsymbol v_{\rm o}\,t+\tfrac12\mathbf a\,t^2 \quad \text{(no difference with that)} \tag{3d-a}\label{3d-a}\\ \boldsymbol v & =\boldsymbol v_{\rm o}+\mathbf a\,t \tag{3d-b}\label{3d-b}\\ \vert\boldsymbol v\vert^2 & =\vert\boldsymbol v_{\rm o}\vert^2+2\,\mathbf a\boldsymbol \cdot\left(\mathbf r-\mathbf r_{\rm o}\right) \tag{3d-c}\label{3d-c} \end{align}

Please explain the general principle of generalization of 1-dimension formulas into 3-dimensions.

And I must add I am very sorry that I print this question not using LaTeX, I really know nothing about it, so I printed it like that, hopefully you will be patient about me.

You can learn LaTex from here: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference — Deep Bhowmik, May 04 '21 at 19:45
Deep's answer is the first intuitive step one can do for 1d to 3 d. Just write them all and add with the corresponding unit vectors and finally club things together and write it as a vector — Sidarth, May 28 '21 at 10:05

Deep Bhowmik · Accepted Answer · 2021-05-28T07:40:36.053

3

Though the above generalisations are correct, what we do in most cases is that we resolute (or break) any given motion along mutually perpendicular axes (namely x, y and z axes) and then apply these formula separately along each of the axes as: $$v_x=v_{0x}+a_{0x}t$$ $$x=v_{0x}t+\frac{1}{2}a_{0x}t^2$$ $$v_x^2=v_{0x}^2+2a_{0x}x$$ along x-axes. Similarly, $$v_y=v_{0y}+a_{0y}t$$ $$y=v_{0y}t+\frac{1}{2}a_{0y}t^2$$ $$v_y^2=v_{0y}^2+2a_{0y}y$$ and $$v_z=v_{0z}+a_{0z}t$$ $$z=v_{0z}t+\frac{1}{2}a_{0z}t^2$$ $$v_z^2=v_{0z}^2+2a_{0z}z$$ along y and z axes respectively. The reason why we do is because motion along mutually perpendicular axes are independent of each other and hence we can apply these formula separately along the axes.

Hope it helps.

edited May 28 '21 at 07:40

answered May 04 '21 at 19:43

Deep Bhowmik

586

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Subscripts for the corresponding acceleration components are needed. Pet-peeve: the acceleration components should also have a 0 for being initial values. – robphy May 05 '21 at 23:02
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These equations are equivalent to OP's equations so isn't the statement that OP's equations are the correct vector generalizations of 1D motion true? – AccidentalTaylorExpansion May 05 '21 at 23:16
@robphy Thanks for correcting the oversight on my part. – Deep Bhowmik May 08 '21 at 13:30
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@AccidentalTaylorExpansion They are true but my style of writing was not appropriate. So I have edited the answer accordingly. – Deep Bhowmik May 08 '21 at 13:31
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For completeness, you might want to include the initial positions, as seen in the OP. – robphy May 08 '21 at 16:36

score 2 · Answer 2 · answered May 04 '21 at 22:58

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As shown in Figure-01

\begin{align} \mathbf r & =\mathbf r_{\rm o}+\boldsymbol v_{\rm o}\,t+\tfrac12\mathbf a\,t^2 \tag{01a}\label{01a}\\ \boldsymbol v & =\boldsymbol v_{\rm o}+\mathbf a\,t \tag{01b}\label{01b}\\ \vert\boldsymbol v\vert^2 & =\vert\boldsymbol v_{\rm o}\vert^2+2\mathbf a\boldsymbol \cdot\left(\mathbf r-\mathbf r_{\rm o}\right) \tag{01c}\label{01c} \end{align}

answered May 04 '21 at 22:58

Frobenius

15,613

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This time, my phone didn't lag while scrolling to the bottom :) – SG8 May 05 '21 at 00:42
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Thanks a lot, I am grateful – Sohaib Ali Alburihy May 05 '21 at 02:54
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@SG8 : Oh ha ha, I use it in my profile as user BioPhysicist post here Special Relativity - Reference Frames S and S′ relative velocity. – Frobenius May 05 '21 at 05:07
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@Sohaib Ali Alburihy : Welcome. Note that Deep Bhowmik's answer is very good and I suggest you to accept it as the best one. – Frobenius May 05 '21 at 05:09
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@Frobenius, I know I saw your profile before but I didn't know that where BioPhysicist stated that :) And, about your answer without words, I found it interesting/useful, so I upvpted whan I saw it. You always bring new perspective of thinking even for the simple/routine problems. This is really good. – SG8 May 05 '21 at 07:28

Generalization of straight line motion under constant acceleration

2 Answers2