If a point mass has some accelerations $\mathbf{a_1} $ and $\mathbf{a_2} $, why is mathematically true that the "total" acceleration is $\mathbf{a}= \mathbf {a_1}+\mathbf {a_2}$?


This is due to the superposition principle: when several forces act upon a body, the net force is the sum of the individual forces: $$\vec F_{net} = \sum \vec F_i $$ However, this is only true when the relation between the force and the acceleration is linear.

Let's take the gravitational force as an example: say you have three bodies and you have already calculated $\vec a_1$ and $\vec a_2$ - the accelerations felt from the third body due to the other two. Then the force on the third one would be $$m \vec a =\vec F_1 + \vec F_2= m \vec a_1 + m \vec a_2 = m(\vec a_1 + \vec a_2)=m\vec a_{1+2} = \vec F_{net} $$ since the force is linear in $\vec a$. Here $\vec a_{1+2}$ - the total acceleration - is really $\vec a_1 + \vec a_2$.

Counter-example: If you had an environment where the acceleration is proportional to the force squared then the superposition principle would not be true. Let's say that this quadratic relationship would be the case for the gravitational force, then the force on the third body would be (I'm just considering the x-component here):

$$\begin{align} m a_x & = (F_{net})^2\\ &=( F_{1x} + F_{2x})^2\\ &=(m a_{1x} + m a_{2x})^2\\ & = (m a_{1x})^2+2m^2 a_{1x} a_{2x}+(m a_{2x} )^2\\ &=( F_{1x})^2+ (F_{2x})^2 + 2m^2 a_{1x} a_{2x}\\ \end{align}$$

The linearity is not given ($(a+b)^2\neq (a^2+b^2)$) and hence the superposition principle not valid. You can see this by looking at the $2m^2 a_{1x}...$ term: in principle the superposition principle just says that the sum of the forces has the same effect as the combination of the individual forces. Although here, the squared sum has the effect of the combined squared forces plus another term.

This in turn means that in this case, the total acceleration which you get on the right hand side is not just $\vec a_1 + \vec a_2$.

| cite | improve this answer | |
  • $\begingroup$ This really answers my question. Superposition principle is a perfectly valid mathematical result :) $\endgroup$ – user132181 Apr 12 '14 at 8:36

It makes no sense for a point mass to have 2 accelerations. What you might have done is find accelerations due to 2 forces separately. You can add them as when $m= \text{constant}$,


When using vectors symbol, its automatically takes care of their directions.

| cite | improve this answer | |
  • $\begingroup$ I can say that $\mathbf {a_1}$ is the acceleration in the $ x $-direction and $\mathbf {a_2} $ is the acceleration in the $ y $-direction, so we can view this particle as having the acceleration $\mathbf {a}=\mathbf {a_1}+\mathbf {a_2 } $. Doesn't that make sense to you? It does to me. But why is it true from the mathematical standpoint? $\endgroup$ – user132181 Apr 8 '14 at 9:16
  • $\begingroup$ Also, your "you can add them" doesn't answer the question. My question is why can you add them? $\endgroup$ – user132181 Apr 8 '14 at 9:23
  • $\begingroup$ @user132181 Have you studied vectors in mathematics? $\endgroup$ – evil999man Apr 8 '14 at 9:24
  • $\begingroup$ @user132181 Vector addition is not like your normal addition. $\endgroup$ – evil999man Apr 8 '14 at 9:24
  • $\begingroup$ Obviously, yes. $\endgroup$ – user132181 Apr 8 '14 at 9:25

While the other answer are all completely correct, I just want to write a more simplified answer.

It's much the same as distances. I you walk 1 meter North and 1 meter East, you can add the two distance vectors and get $\sqrt2$m North-East: $$\vec{d}_1=1m[N]=(1,0),~~\vec d_2=1m[E]=(0,1)$$ $$\vec d=\vec d_1+\vec d_2=(1,1)=1m[N]+1m[E]=\sqrt2m[NE]$$

Adding acceleration vectors works the same way as adding distance vectors. You add the corresponding components (x with x, y with y, etc. whatever coordinates you are using) and the magnitude and direction will work itself out.

| cite | improve this answer | |
  • $\begingroup$ I completely agree on the analogy with velocity (I thought about it too). Why this principle of adding works with accelerations (and possibly jerks, too?) I don't understand, at least from the standpoint of mathematics. $\endgroup$ – user132181 Apr 12 '14 at 6:54
  • $\begingroup$ From the standpoint of mathematics, a vector is a vector. It makes no difference what the units are. $1\frac{m}{s^2}[N]+1\frac{m}{s^2}[E]=\sqrt2\frac{m}{s^2}[NE]$ $\endgroup$ – Jim Apr 12 '14 at 14:05

The expression "Total accelration" does not fit if the accelrations have different directions. The vector resultant is actually the "net accelration", or the combined effect of these two accelrations, or equivalently, forces. The vector resultant makes sure that only the effective components are added, and the opposing effects cancel out.

Maybe an example can help. Consider the following system, where a mass m is acted upon by two accelrations.

Example system

The vector resultant makes sure that the $a\sin \theta$ components are cancelled and the $a\cos \theta$ components are added up. The resultant gives the physically perceived view of motion of the object. A simpler answer would be that accelration is a physical quantity with a direction(i.e. a vector), and when you want to combine two accelrations, you calculate their vector resultant.

| cite | improve this answer | |
  • $\begingroup$ I used the word 'total' loosely, to mean the net acceleration (that's why I just edited the question body and put the word into quotation marks). $\endgroup$ – user132181 Apr 8 '14 at 9:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.