I will first explain how I interpret the vector addition of velocities.

Let's assume that a particle moves with a velocity of $x$ horizontally and simultaneously also moves with a velocity of $y$ vertically.

Now, the resultant velocity will be the vector sum of these individual velocities, i.e. $\vec {v_1} + \vec {v_2}$.

So, in a certain time $t$, the particle experiences a horizontal displacement of $x\cdot t$ and a vertical displacement of $y\cdot t$. So, the resultant displacement is $\sqrt{(xt)^2 + (yt)^2} = \sqrt{t^2(x^2 + y^2)} = t\sqrt{x^2+y^2}$.
Now, the resultant velocity of the path i.e. $\dfrac{\text{Resultant displacement}}{t} = \sqrt{x^2 + y^2}$.

Now, I interpret the vector sum of velocities in terms of displacement because I find it easier to visualize the vector sum of displacements.

Now, how do I interpret the vector sum of acceleration in a way similar to the one I demonstrated above?


  • 3
    $\begingroup$ Since you now understand addition of velocities, can you inductively extend the same argument to acceleration? $\endgroup$
    – Sandejo
    Jul 28, 2020 at 4:57
  • $\begingroup$ @Sandejo Not exactly. That's why I asked this question. I do have a little side question here. Can this understanding be extended to all vectors? Like force, momentum etc.? $\endgroup$ Jul 28, 2020 at 5:09

5 Answers 5


I am not really sure what it is you do not understand or what exactly are you visualizing in your intepretation. But let me just leave this here for you, might be it will help.

You can do more or less the same with acceleration. The problem is however that acceleration can make the motion curved and real distance cannot be computed by pythagorean theorem. This does not happen in a frame in which velocity is initially at rest - then the motion is just straight line and you can write for displacements in vertical and horizontal directions: $$x=\frac{a_x}{2}t^2$$ $$y=\frac{a_y}{2}t^2$$ For diagonal displacement: $$d=\sqrt{x^2+y^2}=\sqrt{\frac{a_x^2+a_y^2}{4}t^4}=\frac{\sqrt{a_x^2+a_y^2}}{2}t^2$$ Comparing this with $d=\frac{a}{2}t^2$ you get $a=\sqrt{a_x^2+a_y^2}.$

Also, in velocity space, the acceleration is speed of how the velocity changes. I.e. from diplacements in velocity space: $$\Delta v_x=a_xt$$ $$\Delta v_y=a_yt$$ You get total velocity displacement traveled $$\Delta v=\sqrt{a_x^2t^2+a_y^2t^2}=t\sqrt{a_x^2+a_y^2}$$ and since acceleration is velocity displacement per time: $$a=\frac{\Delta v}{t}=\sqrt{a_x^2+a_y^2}$$


Starting with the general case

$$x_R=\sqrt{x^2(t)+y^2(t)}\tag 1$$


$$v_R=\dot{x}_R={\frac {x \left( t \right) {\frac {d}{dt}}x \left( t \right) +y \left( t \right) {\frac {d}{dt}}y \left( t \right) }{\sqrt { \left( x \left( t \right) \right) ^{2}+ \left( y \left( t \right) \right) ^{ 2}}}} \tag 2$$

and $$a_R=\ddot{x}_R={\frac { \left( {\frac {d}{dt}}x \left( t \right) \right) ^{2}+x \left( t \right) {\frac {d^{2}}{d{t}^{2}}}x \left( t \right) + \left( {\frac {d}{dt}}y \left( t \right) \right) ^{2}+y \left( t \right) {\frac {d^{2}}{d{t}^{2}}}y \left( t \right) }{\sqrt { \left( x \left( t \right) \right) ^{2}+ \left( y \left( t \right) \right) ^ {2}}}}-1/2\,{\frac { \left( x \left( t \right) {\frac {d}{dt}}x \left( t \right) +y \left( t \right) {\frac {d}{dt}}y \left( t \right) \right) \left( 2\,x \left( t \right) {\frac {d}{dt}}x \left( t \right) +2\,y \left( t \right) {\frac {d}{dt}}y \left( t \right) \right) }{ \left( \left( x \left( t \right) \right) ^{2}+ \left( y \left( t \right) \right) ^{2} \right) ^{3/2}}} \tag 3$$

with $$x(t)=v_x\,t^n~,y(t)=v_y\,t^n$$ you obtain

$$v_R={t}^{n-2}n\sqrt { { \left( x \left( t \right) \right) ^{2}} + \left( y \left( t \right) \right) ^{2}} $$ and

$$a_R=n\,(n-1)\,t^{n-3}\,\sqrt{x^2(t)+y^2(t)} $$

for $n=1$ you get your equation $$v_R\bigg|_{n=1}=\frac{1}{t}\,\sqrt{x^2(t)+y^2(t)}$$ $$v_R\bigg|_{n=2}=2\,t\,\sqrt{x^2(t)+y^2(t)}$$

and $$a_R\bigg|_{n=1}=0$$ $$a_R\bigg|_{n=2}=\frac{2}{t}\,\sqrt{x^2(t)+y^2(t)} $$




I think you mean expressing the resultant acceleration in terms of velocity similar to your approach to the resultant velocity

Let a$_x$ and a$_y$ be the accelerations along x and y axes respectively.

In time t velocity along x changes by a$_x$t and that along y changes by a$_y$t

So from what you have already found net change in v is $\sqrt{{a_xt}^2+{a_yt}^2} $

So net acceleration is (change in v)/t=$\sqrt{{a_x}^2+{a_y}^2} $


Problems with your approach

Your approch might be the most obvious one when starting out, however, it has some major flaws:

  • You assumed constant velocities during the derivation, however, then went on to generalize it for non-constant velocities as well (this flaw can be easily overcome by replacing the $t$ in your equations by $\mathrm dt$, an infinitesimal time)

  • Your derivation, as it is stated, only works for Cartesian coordinates. If you try deriving the velocities using this approach in any other system of coordinates it is highly likely that your approach will fail (especially in curvilinear coordinates)

The second flaw makes your method unreliable and less useful. I advice you to adopt a better method, the one which I present below. I am only primarily showing you how to derive the acceleration, but you can (and you should) also derive the velocity the same way.

Finding acceleration

Acceleration of a particle is defined as

$$\mathbf a=\frac{\mathrm d^2\mathbf r}{\mathrm dt^2}\tag{1}$$

where $\mathbf r$ is the position vector of the particle. Equation $(1)$ is the most general equation we could ever get. Using these equation, you can derive the correct acceleration in any coordinate system, without worrying about any angles, or using trigonometry. Moreover, this approach is fool-proof. It is quite common to not be able to "see" certain components of acceleration (see Coriolis force) when dealing with different coordinate systems, however, using equation $(1)$ you can rest assured and never worry about specific components and their validity.

In Cartesian coordinates, the equation $(1)$ simplifies to

\begin{align} \mathbf a&=\frac{\mathrm d^2(x\mathbf{\hat i}+y\mathbf{\hat j}+z\mathbf{\hat k})}{\mathrm dt^2}\\ &=\frac{\mathrm d^2(x\mathbf{\hat i})}{\mathrm dt^2}+\frac{\mathrm d^2(y\mathbf{\hat j})}{\mathrm dt^2}+\frac{\mathrm d^2(z\mathbf{\hat k})}{\mathrm dt^2}\\ &=\frac{\mathrm d^2 x}{\mathrm dt^2}\mathbf{\hat i}+\frac{\mathrm d^2 y}{\mathrm dt^2}\mathbf{\hat j}+\frac{\mathrm d^2 z}{\mathrm dt^2}\mathbf{\hat k} \end{align}


$$|\mathbf a|=\sqrt{\left(\frac{\mathrm d^2 x}{\mathrm dt^2}\right)^2+\left(\frac{\mathrm d^2 y}{\mathrm dt^2}\right)^2+\left(\frac{\mathrm d^2 z}{\mathrm dt^2}\right)^2}\tag{2}$$

(In the above derivation, I pulled out the unit vectors out of the differentiation, however that might not always be the case. The unit vectors may also change with time, like they do in polar, spherical and cylindrical coordinates)

Like the way I derived the acceleration for Cartesian coordinates, you can also do so for other system of coordinates, such as polar coordinates, spherical coordinates, cylindrical coordinates and oblique coordinates. As a challenge, you could also try deriving them with your method, only to find how difficult it is to derive them using your method, and , instead, how easy it is to derive using the above method.

  • $\begingroup$ "Your derivation, as it is stated, only works for Cartesian coordinates" not true. The approach works just fine for infinitesimal displacements, only the displacement is not given by pythagorean theorem as the triangle has no right angle. $\endgroup$
    – Umaxo
    Aug 5, 2020 at 10:05
  • $\begingroup$ @Umaxo "as it is stated" Of course, with a few modifications, we can do the same for any co-ordinate system, but I prefer to use a universal method instead of figuring out what to do for every different coordinate system. Moreover, since we are using infinitesimal displacements, why not just invoke calculus and do it the better way :-) $\endgroup$
    – user258881
    Aug 5, 2020 at 10:08
  • $\begingroup$ "Of course, with a few modifications" tis but a small inconvenience. The idea itself works as stated, even though I agree that distinction between "idea" and "technical details" might be subjective. $\endgroup$
    – Umaxo
    Aug 5, 2020 at 10:14
  • $\begingroup$ "why not just invoke calculus" The question asks about visualization. $\endgroup$
    – Umaxo
    Aug 5, 2020 at 10:15
  • 2
    $\begingroup$ @Umaxo I do find calculus visual, but then again, the OP might not resonate with me. And as far as the OP's method goes, I find it less powerful. It's okay to get a feel for some basic/tradituonal scenarios, but primarily, the math should be the guiding light. $\endgroup$
    – user258881
    Aug 5, 2020 at 10:24

You can give the answer your own way by taking a situation like: A particle originally at rest has acceleration of $x$ $m$/$sec^2$ horizontally and $y$ $m$/$sec^2$ vertically.

Now the resultant acceleration would be the vector sum of the individual accelerations, i.e. $\vec {a_1} + \vec {a_2}$.

After sometime velocity along X-direction and along Y-direction would be $xt$ $m$/$sec$ and $yt$ $m$/$sec$ respectively.

Now, in accordance with you the resultant velocity would be $\sqrt{(xt)^2 + (yt)^2} = \sqrt{t^2(x^2 + y^2)} = t\sqrt{x^2+y^2}$.

The acceleration of the particle is $\dfrac{\text{Change in velocity}}{time} = \sqrt{x^2 + y^2}$ $m$/$sec^2$.

You can use this understanding for any type of vector addition.


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