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Could I use $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ for calculating velocity in these 3 different dimensions? If not, what's the difference between these 3 dimensions? How would you calculate velocity in 3d? One more thing, is $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ for 3d, 2d or both?

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As long as you know what it means to add vectors and to multiple a vector by a scalar these equations remain the same in any dimension. That said a lot of students get stuck on how to handle vectors for a while. You have not made it clear if you know the difference between vectors and scalars and how comfortable you are working with them. –  dmckee Feb 8 at 18:50
    
physics has been giving me a hard time and i've been trying to learn it. i was just wondering about this because i've this physics teaching game idea (when i'm done learning physics) and i want it to be in 3d but want to know what dimension of physics that'll be used. –  Physicsnoob Feb 8 at 19:00

3 Answers 3

In terms of velocitiy in classical mechanics the only difference between the dimensions is what kind of objects $v$, $u$ and $a$ are.

While $t$ should always be a real number $v$,$u$ and $a$ should be vectors in $\mathbb{R}^n$ where $n$ represents the dimension you are talking about. If this does not help, please clarify.

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it doesn't help. –  Physicsnoob Feb 8 at 18:45
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@Physicsnoob could you clarify why it doesn't help? Is there something you don't understand? –  David Z Feb 8 at 19:35

Notice that in more than one dimension (two, three, and even higher!) the boldface vector notation $\mathbf x(t)$ is really just shorthand for a tuple of component functions. Explicitly, in dimension $d$, one has \begin{align} \mathbf x(t) = \begin{pmatrix} x^1(t) \\ \vdots \\ x^d(t) \\ \end{pmatrix} \end{align} Now, the velocity vector is defined as the derivative of $\mathbf x$ with respect to time; $\mathbf v(t) = \dot{\mathbf x}(t)$ where the overdot conventionally means a derivative with respect to time. One can show that this implies that the velocity vector can be obtained by simply differentiating the position vector components; \begin{align} \mathbf v(t) = \begin{pmatrix} v^1(t) \\ \vdots \\ v^d(t) \\ \end{pmatrix} = \begin{pmatrix} \dot x^1(t) \\ \vdots \\ \dot x^d(t) \\ \end{pmatrix} \end{align} Similarly, the acceleration is defined as the derivative of the velocity with respect to time, and it's components are simply the first derivatives of the velocity components with respect to time or, equivalently, the second derivatives of the position components with respect to time; \begin{align} \mathbf a(t) = \begin{pmatrix} a^1(t) \\ \vdots \\ a^d(t) \\ \end{pmatrix} = \begin{pmatrix} \dot v^1(t) \\ \vdots \\ \dot v^d(t) \\ \end{pmatrix} = \begin{pmatrix} \ddot x^1(t) \\ \vdots \\ \ddot x^d(t) \\ \end{pmatrix} \end{align} Now, suppose that the acceleration is constant, namely $\mathbf a(t) = \mathbf a_0$ for some constant $\mathbf a_0$. In boldface notation, this could be written in terms of the velocity as $\dot{\mathbf v}(t) = \mathbf a_0$. On the other hand, in component form, it would be written as \begin{align} \begin{pmatrix} \dot v^1(t) \\ \vdots \\ \dot v^d(t) \\ \end{pmatrix} = \begin{pmatrix} (a_0)^1 \\ \vdots \\ (a_0)^d \\ \end{pmatrix} \end{align} but recall that two such $d$-tuples are equal if and only if their components are equal, so this equality of vectors leads to a set of $d$, $1$-dimensional equations, one for each component; \begin{align} \dot v^i(t) = (a_0)^i(t). \end{align} Now each of these is a one-dimensional equation whose solution is $ v^i(t) = v^i(0) + (a_0)^i t$ for each $i=1, \dots d$. But this is equivalent to \begin{align} \begin{pmatrix} v^1(t) \\ \vdots \\ v^d(t) \\ \end{pmatrix} = \begin{pmatrix} v^1(0) + (a_0)^1t \\ \vdots \\ v^d(0) + (a_0)^dt \\ \end{pmatrix} \end{align} which in boldface notation is equivalent to \begin{align} \mathbf v(t) = \mathbf v(0) + \mathbf a_0 t \end{align} as desired.

Note. This works in any dimension $d\geq 1$, so it works in dimensions $2$ and $3$ as special cases.

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One dimensional motion

  • The motion of an object is called one dimensional, if only one of the three co-ordinates required to specify the position of the object in space changes w.r.t time. In such a motion, the object moves along a straight line. For example, motion of a train along a straight railway track, a man walking on a level and narrow road, an object falling freely under gravity, etc.
    enter image description here
  • Velocity of an object is given by different equations depending on the situation.

    $1.$ When the velocity of object is constant, velocity will be same at any given time. In that case velocity is given by the following equation,
    $$\overrightarrow{v}=\overrightarrow{v_x}=\frac{\overrightarrow{x_2}-\overrightarrow{x_1}}{t_2-t_1}=\frac{\triangle{\overrightarrow{x}}}{\triangle{t}}$$
    $2.$ When the same change in velocity of object takes place in each unit of time (uniformly accelerated motion) velocity of the object at time $t_1$ and velocity of the object at time $t_2$ can be given by the following equation, $$\overrightarrow{v_2}=\overrightarrow{v_1}+\overrightarrow{a}(t_2-t_1) ........(ii)$$ For one dimensional motion the above equation reduces to the following form,
    $$\overrightarrow{v_{2x}}=\overrightarrow{v_{1x}}+\overrightarrow{a_x}(t_2-t_1)$$
    Magnitude can be given by the following equation,
    $$v_{2x}=v_{1x}+a_x(t_2-t_1)$$
    In simple form we write the above equation as following,
    $$v^{'}=v+a(t^{'}-t)$$

    $3.$ When the object is moving with variable velocity, it undergoes unequal displacements in equal intervals of time.
    In that case velocity at the particular time can be found by using differentiation. If the time interval around the instant at which velocity is to be determined is made small and small, then in the limiting case when $\triangle{t}$ tends to zero, the value of $\frac{\triangle{x}}{\triangle{t}}$ approaches the velocity of the object at that time. Mathematically the instantaneous velocity is given by the following equation,
    $$\overrightarrow{v}=Lt_{\triangle{t}\rightarrow0}(\frac{\triangle{\overrightarrow{x}}}{\triangle{t}})$$ In the language of calculus, $$Lt_{\triangle{t}\rightarrow0}(\frac{\triangle{\overrightarrow{x}}}{\triangle{t}})=\frac{d\overrightarrow{x}}{dt}$$

Two dimensional motion

  • The motion of an object is called two dimensional, if two of the three coordinates required to specify the position of the object in space change w.r.t time.In such a motion, the object moves in a plane. For example, a billiard ball moving over the billiard table, an insect crawling over the floor of a room, etc.
    enter image description here
  • Velocity of an object is given by different equations depending on the situation.

    $1.$ When the velocity of object is constant, velocity will be same at any given time. In that case velocity is given by the following equation,
    $$\overrightarrow{v}=\overrightarrow{v_x}+\overrightarrow{v_y}=\frac{\overrightarrow{x_2}-\overrightarrow{x_1}}{t_2-t_1}+\frac{\overrightarrow{y_2}-\overrightarrow{y_1}}{t_2-t_1}=\frac{\triangle{\overrightarrow{x}}}{\triangle{t}}+\frac{\triangle{\overrightarrow{y}}}{\triangle{t}}$$
    $2.$ When the same change in velocity of object takes place in each unit of time (uniformly accelerated motion) velocity of the object at time $t_1$ and velocity of the object at time $t_2$ can be given by the following equation, $$\overrightarrow{v_2}=\overrightarrow{v_1}+\overrightarrow{a}(t_2-t_1) ........(ii)$$ For two dimensional motion the above equation reduces to the following form,
    $$(\overrightarrow{v_{2x}}+\overrightarrow{v_{2y}})=(\overrightarrow{v_{1x}}+\overrightarrow{v_{1y}})+(\overrightarrow{a_x}+\overrightarrow{a_y})(t_2-t_1)$$
    Magnitude can be given by the following equation,
    $$(\sqrt{v_{2x}^2+v_{2y}^2})=(\sqrt{v_{1x}^2+v_{1y}^2})+(\sqrt{a_x^2+a_y^2})(t_2-t_1)$$
    $3.$ When the object is moving with variable velocity, velocity in two dimensional motion can be given as follows,
    $$\overrightarrow{v}=Lt_{\triangle{t}\rightarrow0}(\frac{\triangle{\overrightarrow{x}}}{\triangle{t}})+Lt_{\triangle{t}\rightarrow0}(\frac{\triangle{\overrightarrow{y}}}{\triangle{t}})=\frac{d\overrightarrow{x}}{dt}+\frac{d\overrightarrow{y}}{dt}$$

Three dimensional motion

  • The motion of an object is called three dimensional, if all the three coordinates required to specify the position of the object in space change w.r.t time. A few examples of three dimensional motion are a flying kite, a flying bird, etc.
    enter image description here
  • Velocity of an object is given by different equations depending on the situation.

    $1.$ When the velocity of object is constant, velocity will be same at any given time. In that case velocity is given by the following equation,
    $$\overrightarrow{v}=\overrightarrow{v_x}+\overrightarrow{v_y}+\overrightarrow{v_y}=\frac{\overrightarrow{x_2}-\overrightarrow{x_1}}{t_2-t_1}+\frac{\overrightarrow{y_2}-\overrightarrow{y_1}}{t_2-t_1}+\frac{\overrightarrow{z_2}-\overrightarrow{z_1}}{t_2-t_1}=\frac{\triangle{\overrightarrow{x}}}{\triangle{t}}+\frac{\triangle{\overrightarrow{y}}}{\triangle{t}}+\frac{\triangle{\overrightarrow{z}}}{\triangle{t}}$$
    $2.$ When the same change in velocity of object takes place in each unit of time (uniformly accelerated motion) velocity of the object at time $t_1$ and velocity of the object at time $t_2$ can be given by the following equation, $$\overrightarrow{v_2}=\overrightarrow{v_1}+\overrightarrow{a}(t_2-t_1) ........(ii)$$ For three dimensional motion the above equation reduces to the following form,
    $$(\overrightarrow{v_{2x}}+\overrightarrow{v_{2y}}+\overrightarrow{v_{2z}})=(\overrightarrow{v_{1x}}+\overrightarrow{v_{1y}}+\overrightarrow{v_{1z}})+(\overrightarrow{a_x}+\overrightarrow{a_y}+\overrightarrow{a_z})(t_2-t_1)$$
    Magnitude can be given by the following equation,
    $$(\sqrt{v_{2x}^2+v_{2y}^2+v_{2z}^2})=(\sqrt{v_{1x}^2+v_{1y}^2+v_{1z}^2})+(\sqrt{a_x^2+a_y^2+a_z^2})(t_2-t_1)$$
    $3.$ When the object is moving with variable velocity, velocity in three dimensional motion can be given as follows,
    $$\overrightarrow{v}=Lt_{\triangle{t}\rightarrow0}(\frac{\triangle{\overrightarrow{x}}}{\triangle{t}})+Lt_{\triangle{t}\rightarrow0}(\frac{\triangle{\overrightarrow{y}}}{\triangle{t}})+Lt_{\triangle{t}\rightarrow0}(\frac{\triangle{\overrightarrow{z}}}{\triangle{t}})=\frac{d\overrightarrow{x}}{dt}+\frac{d\overrightarrow{y}}{dt}+\frac{d\overrightarrow{z}}{dt}$$

I hope this helped.

(Any correction advisory is welcome)

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It did. I just really wanted to know since the game will be teaching physics and i needed to know whether the dimension of the game will affect what physics it'll be teaching. I'll be using the game objects to teach the physics, does this affect anything. –  Physicsnoob Feb 8 at 21:34
    
Suppose, we take football (3D) as the object for teaching. We can specify the position of football in space by set of three mutually perpendicular lines, called the co-ordinate frame. If we prefer to teach 1D motion, we move the football in a straight line (say, in the direction of X-axis), the change in the position of football will be only along X-axis, thus change in other two coordinates will be zero. Similarly, we can use 3D objects to teach 2D and 3D motion by varying 2 or 3 coordinates. So, using 3D objects in teaching 1D,2D,3D motion doesn't affect. –  Godparticle Feb 8 at 22:22
    
ok. What of other physics formula/equations? This was the formula i saw on a site: dx/something + dy/something and they said if it was 3d, you just add dz/over. I can't remember what it was divided by –  Physicsnoob Feb 8 at 23:13
    
@Physicsnoob What you are referring to is calculus but I dont know the exact formula you saw. You should have a good understanding of the basic formulas before you start delving into calculus. –  stackErr Feb 9 at 0:03
    
@Physicsnoob. Would you please link that site about which you are speaking about. I am not getting what you are saying. –  Godparticle Feb 9 at 2:38

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