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I have been learning about vectors and acceleration recently and I still don't understand how to determine the direction of a vector.

For instance, if we consider a freely falling particle and decide that the positive y-axis is going to be "up" then the acceleration should be negative; however I thought this was because the acceleration is increasing as we go "down," but if this was the reason then the velocity should also be negative because it is increasing as we go down... Still, I think the velocity should be positive, is that correct?

So my question is, how do I determine the direction of velocity, displacement and acceleration? And more generally, how do I determine the direction of any vector, so that I don't have to ask again when I have a different quantity? Are there any rules?

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  • $\begingroup$ The velocity changes direction going from up to down. Acceleration in your example is always negative. By the way it is not increasing on the way up, it is decreasing. It is also decreasing on the way down since -10 is less than -5. $\endgroup$ – ggcg Dec 29 '18 at 0:34
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Sure, there are rules, and it is pretty simple. You'll see after you solve this confusion.

Suppose you come up with a kinematics exercise. After you read and understand it (yes, it's obvious that's the first thing, but not everybody does haha), then we must set a reference frame.

Setting the reference frame means not only deciding "the reference point" from which we will measure distances. It also includes deciding the axes.

So, for example, imagine that you read the problem and it is a free fall, because you read that it is a free fall. An intelligent choice would be setting the reference frame (RF) at the bottom, and the usual cartesian axes. Doing it like this, distances will be possitive, and that's good.

So we've chosen two perpendicular axes. The usual cartesian axes. You've chosen where to put the origin of those axes (preferably at the bottom). So now we've got the axes. From now on:

  • Positions are given by the coordinates of the axes.

If you've chosen the usual cartesian axes, then it is possitive above the origin and negative below it. (Also possitive at the right of the origin and negative at the left). This has been like this from the beginning of times.


Now, velocities.

You have to keep in mind the definition:

$$v=\frac{\Delta s}{\Delta t}$$

Where $\Delta$ means "increase". So

$v$= how much distances increases $\div$ how much time increases.

But time flows always forward, never backwards. So the increasing of time is always possitive, whatever its value. $\Delta t>0$. This means that the sign of velocity is given by the change in distance:

  • If distance increases, $v$ is possitive.
  • If distance decreases, $v$ is negative.

This is usually possitive if it goes upwards/rightwards, and negative if it is downwards/leftwards.

But this is just because of the definition. If the distance is decreasing (that means everytime less possitive, or more negative), then $\Delta s<0$ and thus $v<0$.


For the same reason, accelerations are

$$a=\frac{\Delta v}{\Delta t}$$

So the sign of acceleration is given by the sign of $\Delta v$, that is, the sign of the "change of $v$".

If $v$ is decreasing, acceleration is negative. If $v$ is increasing, acceleration is possitive.

Check that we are usually dealing with negative numebrs too. So "Increasing" can be regarded as "more possitive" or "less negative", equivalently.


Back to the exercise.

So, the particular case of a free fall. We are working on the vertical axis, labeled $y$.

The intelligent choice is setting the origin at the bottom. Like this, your initial position would be $0$ if it starts from the ground, or a possitive numebr if it starts from certain height, but possiive anyways, and that's good.

$y_0>0$.

Now, velocity. It depends on the problem. If the object is initially moving upwards, $v_0>0$. If it is moving downwards, $v<0$.

Acceleration is certainly negative, because it tends to decrease velocity. If $v$ was possitive, it is trying to stop the particle. IF $v$ was already negative, acceleration will make it go downwards faster, so velocity is smaller, in the sense of "more absolute value but negative". It's "more negative", so it's still decreasing. Acceleration is negative anyways.

But this is because of your choice of axis. IF you had chosen a "reversed axis", so that possitive coordinates were below the origin, acceleration would be possitive. It's all about thinking with the definitions: is velocity making position grow or decrease?

And we deal with real numbers. Going from minus 10 to minus 20 is decreasing 10 units. It's bigger in absolute value, but the sign is there.

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  • $\begingroup$ Whoa, I tried to make it complete, but re-reading it, I find it messy. Did you understand? You can ask for clarification. $\endgroup$ – FGSUZ Dec 29 '18 at 0:01
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Of course there are rules.

Step 1: Define your coordinate system. That is, choose a set of perpendicular (conventionally right handed) $x,y,z$ axes, and which directions are positive. This must come first.

Step 2: STICK to the conventions you defined in step 1, and apply all the usual rules of vectors.

Example: Define a system in which we take $+y$ to be down and $+x$ to the left. Now suppose that you throw a ball to the $right$ off a cliff (from the origin) with no angle upward (so just straight off). Its coordinates are as always given by the vector $\mathbf{r} = (x,y)$.

If we actually plug in numbers now (say at a certain time), we see that the ball's position will look something like $\mathbf{r}=(-1, 1)$, since it is lower than where it started (positive y) and to the right of where we started (-x).

The velocity is always be given by $\mathbf{v} = (\dot{x}, \dot{y})$ where the dot denotes the time derivative. A quick sketch will show you that $\dot{x}$ is negative and $\dot{y}$ will be positive.

The acceleration as always is given by $\mathbf{a} = (\ddot{x}, \ddot{y})$. In our example, $\ddot{x}=0$ and $\ddot{y} = + g$.

And so we have completely described the system.

When in doubt, draw a picture of the problem with the conventions you defined in step 1 and resolve into components to check your signs.

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Let $\bf g$ be the acceleration due to gravity. It is a vector. Let $\bf v$ be the velocity of some object (also a vector) and let $\bf a$ be the acceleration of that object. If the object is subject only to gravity, then ${\bf a} = {\bf g}$.

Let's say the $y$ axis has been set up to point in the upwards direction. In this case the acceleration due to gravity is ${\bf g} = (0,-g,0)$. In other words, it has no $x$ and $z$ component, and its $y$ component is $-g$ where $g \simeq 9.8$ m/s$^2$.

In the situation under discussion, an object moving in the upwards direction has a velocity with positive $y$ component and an object moving in the downwards direction has a velocity with negative $y$ component.

Let $v_y$ be the $y$-component of the velocity. An object thrown upwards will start with $v_y$ positive. As time goes on, $v_y$ will fall, reaching zero at the top of the trajectory. After that $v_y$ will be negative, and get more and more negative as the object accelerates in the downwards direction.

As for the vector $\bf v$, it is pointing in the upwards (positive $y$) direction when $v_y > 0$ and it is pointing in the downwards (negative $y$) direction when $v_y < 0$.

Finally, you ask if there are rules about the direction of vectors. The answer is mainly "don't worry, you will understand this more as you work with vectors more". Your question is a bit like asking if there are rules for the direction of arrows. In physics most vectors can point in pretty much any direction, depending on the physical situation, but once you have the initial direction of a given vector, its direction after that is determined by whatever processes (forces etc.) are going on. The example of a thrown object illustrates this. You could throw an object either in the upwards or the downwards direction, but after you let go the local gravity will determine what happens to the velocity in the subsequent motion.

When setting up a physical problem, it is often useful to draw a diagram with arrows representing vectors. This could be done for the forces acting on a body. If you know at the start what direction all the forces are acting in, then you can draw the arrows accordingly. However, when working with velocity it often happens that we don't know what direction the velocity is going to be in. In this case it is best to draw the arrow in the positive direction, and write equations accordingly. If the answer comes out that the velocity is negative, then you know it means it is in the opposite direction to the one you chose to call positive when you drew the arrow at the start of the calculation. You will very likely not grasp this right away from only reading this answer, but you will grasp it after putting in practice on a lot of examples!

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Let $\hat u$ be the unit vector in the up direction.
A vector $\vec a$ can be written as $a\, \hat u$ where $a$ is the component of the vector in the up ($\hat u$) direction and $a$ can be a positive or negative quantity.
The magnitude (or length) of the vector is $|a|$.

In your example let the acceleration of free fall be $-10 \, \hat u$.
The component of the acceleration in the $\hat u$ direction is $-10$ and the magnitude of the acceleration is $|-1| = 10$
Another way of writing the acceleration is $10\,(-\hat yu)$ ie the acceleration has a component of $10$ in the $(-\hat u)$ or down direction with the magnitude still $10$.

Suppose you drop a body from rest and want to find its velocity after a time of $3$ seconds using the constant acceleration kinematic equation $\vec v_{\rm final} = \vec v_{\rm initial} + t \, \vec a$.

In this case $v_{\rm initial} = 0\, \hat u,\,a=-10\,\hat u$ and $t=3$.

$\vec v_{\rm final} = v_{\rm final} \,\hat u = 0\,\hat u + 3\,(-10 \hat u) = -30 \,\hat u \Rightarrow v_{\rm final} = -30$

So the final velocity has a component of $-30$ in the $\hat u$ direction.

Put another way the final velocity has a magnitude of $|-30|=30$ and is in the $(-\hat u)$ or down direction.

Which ever way you look at it the speed (magnitude of velocity) of the body has increased by $30$.

Note that another way of writing the final velocity is $30\, (-\hat u)$ ie the velocity has a component of $30$ in the $(-\hat u)$ or down direction.

Whichever way you choose the write the vector the important thing as far as your question is concerned is that the magnitude of the velocity (or speed) of the body has increased form $0$ to $30$ in three seconds.


When doing such a problem what one tends to do is to write an equation which links the components of the various vectors involved.
So the constant acceleration kinematic equation is written as $v_{\rm final} = v_{\rm initial} + a t$ and substitute the components of the vectors in the up $(\hat u)$ direction $v_{\rm initial} =0$ and $g = -10$ into the equation to find the component of the final velocity in the up $(\hat u)$ direction which in this case turns out to be $v_{\rm final} = 0 + (-10) \, 3 = -30$.

This answer you interpret as the velocity having a magnitude of $|-30| =30$ in the downward $(-\hat u)$direction.


If you chose down as positive with unit vector $\hat d$ then you would find that $\vec v_{\rm final} = 30\,\hat d$ or $v_{\rm final} = 30$


As there is a simple connection between the two unit vectors in that $\hat d = -1\, \hat u$ or $\hat d = (-\hat u$) it does not matter which direction you choose as the positive direction, just be consistent.

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You have every right to be confused. Vectors can be taught in a confusing way.

The basic vector is displacement. If you go from A to B your displacement, $\vec{AB},$ is in the direction from A to B. That's it!

Velocity is probably the next easiest vector. If you go from A to B, in a time $\Delta t, $ your mean velocity is defined by:$$\text{mean velocity}=\frac{\text{displacement } \vec{AB}}{\Delta t}$$ This implies that $\vec{AB}$ is the direction of the average velocity. Crystal clear so far, I hope. Instantaneous velocity is defined as the limiting value of the mean velocity as we make the time interval smaller and smaller, centring the interval on the instant at which we require the velocity.

The mean acceleration over a time interval $\Delta t$ is defined by$$\text{mean acceleration}=\frac{\text{velocity at end of} \Delta t-\text{velocity at start of} \Delta t}{\Delta t}$$

According to this definition, the acceleration due to gravity is downwards, because the top line of the fraction is a vector in the downward direction. Even if we throw a ball upwards, the acceleration is downwards, because the pull of the Earth makes the body lose velocity upwards, which is equivalent to its gaining velocity downwards.

Like all geometrically-based vectors, acceleration has a magnitude and a direction. It is not positive or negative!

But… It is often convenient to employ "unit vectors", vectors of magnitude 1 pointing in chosen directions. Suppose we choose the upward direction. In that case the acceleration due to gravity can be written as $-9.81 \text{m s}^{-2} \times\ \text{upward unit vector}.$ Another correct way to say this is that the upward component of the acceleration due to gravity is $-9.81 \text{m s}^{-2}.$ If you choose a downward unit vector, then the acceleration is $+9.81 \text{m s}^{-2} \times\ \text{downward unit vector}.$ In other words the downward component of the acceleration is $+9.81 \text{m s}^{-2}.$ Writers of elementary textbooks seldom mention unit vectors or components in this context.

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