# Negative Velocity or Acceleration

So, velocity is a vector, right?
And vectors can't have negative magnitudes, right?
Then why is leftward velocity considered negative in 1D kinematics? It just seems off to me.
Same with acceleration, and pretty much _every vector in all of physics._

Ibix
So, velocity is a vector, right?
And vectors can't have negative magnitudes, right?
Yes and yes.
Then why is leftward velocity considered negative in 1D kinematics?
The components of a vector can be negative. The magnitude is the (positive) square root of the sum of the squares of the components. Which means that the magnitude of your 1d vector, with its one component, is ##\sqrt{v_x^2}##, which is positive.
It just seems off to me.
Same with acceleration, and pretty much _every vector in all of physics._
How would you represent velocities in the opposite direction if you didn't use signed numbers?

FactChecker
BvU
Homework Helper
So, velocity is a vector, right?
Yes
And vectors can't have negative magnitudes, right?
Yes
Then why is leftward velocity considered negative in 1D kinematics?
Because it is a (cartesian) component of a vector, not a vector.
It just seems off to me.
If you stand next to a canal with the canal on your right hand side and jump 1 m sideways you can either stay dry or get very wet. But -1 m to the right and you stay dry. There is a difference between 'how far you jump' (always non-negative) and 'how far you jump in a given direction'.
Same with acceleration, and pretty much _every vector in all of physics._
Yes. And very useful. If you take two jumps of 1 m each you can end up at the point you started. Apparently you can not simply add magnitudes, but you can add vectors.

Yes and yes.
The components of a vector can be negative. The magnitude is the (positive) square root of the sum of the squares of the components. Which means that the magnitude of your 1d vector, with its one component, is ##\sqrt{v_x^2}##, which is positive.
How would you represent velocities in the opposite direction if you didn't use signed numbers?
Yes
Yes
Because it is a (cartesian) component of a vector, not a vector.
If you stand next to a canal with the canal on your right hand side and jump 1 m sideways you can either stay dry or get very wet. But -1 m to the right and you stay dry. There is a difference between 'how far you jump' (always non-negative) and 'how far you jump in a given direction'.
Yes. And very useful. If you take two jumps of 1 m each you can end up at the point you started. Apparently you can not simply add magnitudes, but you can add vectors.
So, in 2D kinematics, velocities must have two signs, one for each component. (Translation: I still don't understand.)

Nugatory
Mentor
So, in 2D kinematics, velocities must have two signs, one for each component. (Translation: I still don't understand.)
If I'm moving northwards, the NS component of my velocity is positive and the EW component is zero. If I'm moving westwards, the EW component is negative and the NS component is zero. If the NS component and the EW component are both positive, then I'm moving generally towards the northeast; but if the NS component is negative and the EW component is positive I'm moving generally towards the southeast.

jbriggs444
Homework Helper
So, in 2D kinematics, velocities must have two signs, one for each component. (Translation: I still don't understand.)
The relevant concept is not "sign", it is "direction".

A "direction vector" (also known as "unit vector") is a vector whose magnitude is 1. Any vector can be expressed as the product of a direction vector and a non-negative scalar.

For any vector (other than the zero vector), the decomposition into a direction and a scalar magnitude is unique. There is only one direction associated with the vector.

In one dimension, there are two direction vectors: Expressed as 1-tuples, they are (1) and (-1). In this special case, the notions of "sign" and of "direction" coincide.

In two dimensions, there are infinitely many direction vectors. Graphically, they form a circle with radius 1 centered at the origin. Expressed as 2-tuples, they are (x,y) where ##\sqrt{x^2+y^2} = 1##

In three dimensions, there are infinitely many direction vectors. Graphically, they form a spherical shell with radius 1 centered at the origin. Expressed as 3-tuples, they are (x,y,z) where ##\sqrt{x^2+y^2+z^2} = 1##

BvU and enter
The relevant concept is not "sign", it is "direction".

A "direction vector" (also known as "unit vector") is a vector whose magnitude is 1. Any vector can be expressed as the product of a direction vector and a non-negative scalar.

For any vector (other than the zero vector), the decomposition into a direction and a scalar magnitude is unique. There is only one direction associated with the vector.

In one dimension, there are two direction vectors: Expressed as 1-tuples, they are (1) and (-1). In this special case, the notions of "sign" and of "direction" coincide.

In two dimensions, there are infinitely many direction vectors. Graphically, they form a circle with radius 1 centered at the origin. Expressed as 2-tuples, they are (x,y) where ##\sqrt{x^2+y^2} = 1##

In three dimensions, there are infinitely many direction vectors. Graphically, they form a spherical shell with radius 1 centered at the origin. Expressed as 3-tuples, they are (x,y,z) where ##\sqrt{x^2+y^2+z^2} = 1##
Ohh, now it clicked! Thanks! I got confused by what Ibix said and understood it as sign and not direction.

By the way, when would we say that a velocity is negative in _2D space_?

Chestermiller
Mentor
To elaborate on what jbriggs444 said, if the velocity vector or the acceleration vector are pointing to the left (in the negative x-direction), then we represent these vectors as $$\mathbf{v}=v(-\mathbf{i_x})$$ and $$\mathbf{a}=a(-\mathbf{i_x})$$where v is the magnitude of the velocity vector and a is the magnitude of the acceleration vector. In this representation, the magnitudes are positive, but the directions are negative. When our vector equations (e.g., vectorial force balance equation) are dotted with the unit vector in the positive x-direction (##\mathbf{i_x}##), the negative sign remains.

CWatters
Homework Helper
Gold Member
So, in 2D kinematics, velocities must have two signs, one for each component. (Translation: I still don't understand.)

Yes (if you use x,y co-ordinates to specify the direction).

The direction component can be specified many different ways, for example a ships navigator might refer to a vector as "3 miles north west". Using that co-ordinate system the direction component doesn't have a sign.

In the 1D example you could use Left and Right to define the direction of your vectors and then they wouldn't have a sign either. Personally I think that would make the maths harder.

A sum like
3 + (-4) - (-7) = ?

becomes
3 Right + 4 Left - 7 Left = ?

jbriggs444
Homework Helper
By the way, when would we say that a velocity is negative in _2D space_?
I would never say that a velocity is negative in 2D space. I would only say that a velocity is negative if the 2D space were somehow reduced to a 1D space.

A few examples:

If a fly is buzzing around a pasture, I would not refer to its velocity as either positive or negative.

If a car is driving down a road I could note the direction the car is pointing and use that to impute "forward" and "backward" directions, associate positive with forward and negative with backward and refer to its the velocity as positive or negative. In this case the 2D space has been reduced to a 1D space.

If the same car were sliding sideways on an icy lake, I would not refer to its velocity as either positive or negative.

jbriggs444
Homework Helper
Yes (if you use x,y co-ordinates to specify the direction).

The direction component can be specified many different ways, for example a ships navigator might refer to a vector as "3 miles north west". Using that co-ordinate system the direction component doesn't have a sign.
Of course using coordinates means that one needs a coordinate system to nail down the sign conventions.

In one dimension there are only two possible sign conventions. In two dimensions, there are infinitely many "two sign" conventions. One for every choice of rotation angle for the coordinate system. [Times two if one is picky].

I would never say that a velocity is negative in 2D space. I would only say that a velocity is negative if the 2D space were somehow reduced to a 1D space.

A few examples:

If a fly is buzzing around a pasture, I would not refer to its velocity as either positive or negative.

If a car is driving down a road I could note the direction the car is pointing and use that to impute "forward" and "backward" directions, associate positive with forward and negative with backward and refer to its the velocity as positive or negative. In this case the 2D space has been reduced to a 1D space.

If the same car were sliding sideways on an icy lake, I would not refer to its velocity as either positive or negative.
The reason that I asked this was that I got 4-5 wrong answers while practicing 1D kinematics, all of them about negative velocities. Should I worry about negativity in velocities while practicing/applying 2D/3D kinematics? From what I understood from your reply is that I shouldn't (except for reduction to 1D space), but I'm not sure so I wanted to clarify.

P.S. Wow, this place is more active than I thought.

jbriggs444
Homework Helper
The reason that I asked this was that I got 4-5 wrong answers while practicing 1D kinematics, all of them about negative velocities. Should I worry about negativity in velocities while practicing/applying 2D/3D kinematics? From what I understood from your reply is that I shouldn't (except for reduction to 1D space), but I'm not sure so I wanted to clarify.
Personally, I got through introductory physics without worrying much about sign conventions and allowed intuition do its work. It always seemed clear in which direction various effects would act and I could adjust signs as needed to obtain the correct result.

In retrospect, that was not a good way to proceed.

Being able to fall back on a formal methodology to get the signs right is useful. @Chestermiller can give you far better advice than I in that regard.

Last edited:
CWatters
Homework Helper
Gold Member
The reason that I asked this was that I got 4-5 wrong answers while practicing 1D kinematics, all of them about negative velocities. Should I worry about negativity in velocities while practicing/applying 2D/3D kinematics? From what I understood from your reply is that I shouldn't (except for reduction to 1D space), but I'm not sure so I wanted to clarify.

P.S. Wow, this place is more active than I thought.

Its usually important to keep track of signs because they tell you about the direction.

In projectile motion problems (throwing or dropping rocks or firing cannons) it's good practice to define what you mean by positive. For example suppose the problem involves throwing a rock upwards at 45 degrees from the top of a cliff. Your answer might start with...

"I define the origin to be the top of the cliff, and up and right to be positive"

Then later if the answer has a positive horizontal component of velocity you know the rock is still moving to the right. If the vertical component of velocity is negative then you know it has stopped going up and is on the way down.

In some cases you have to solve a quadratic equation and the answer may have two values. For example that rock will be 2 meters above the launch point twice, once on the way up and once on the way down. A bit later it will be at -2m. Normally it's obvious which is the right answer but sometimes byou need to look at the sign of the velocity to work out which is the right point.

jbriggs444
A.T.
The relevant concept is not "sign", it is "direction".
It's important to keep these two apart, because sometimes the sign convention is not based on a coordinate system, but a diagram, in which (eventually opposite arrows) indicate the positive direction for each vector individually.

PeterO
Homework Helper
So, velocity is a vector, right?
And vectors can't have negative magnitudes, right?
Then why is leftward velocity considered negative in 1D kinematics? It just seems off to me.
Same with acceleration, and pretty much _every vector in all of physics._
I am not sure why you think leftward velocity is negative. The choice of right or left for positive is completely up to you. Indeed, if you were analysing a situation where all motion is directed to the left, I would certainly define left as positive - only to remove most of the negative values from my work.
It is the same with vertical motion. When analysing a case when a ball is thrown vertically up, and it eventually comes back down again, most people will define UP as positive and down as negative. When analysing a case where a rock is dropped from a bridge and falls into a river, we generally define DOWN as positive.

BvU
Homework Helper
The reason that I asked this was that I got 4-5 wrong answers while practicing 1D kinematics, all of them about negative velocities. Should I worry about negativity in velocities while practicing/applying 2D/3D kinematics? From what I understood from your reply is that I shouldn't (except for reduction to 1D space), but I'm not sure so I wanted to clarify.
You should, yes.

Take the simple example F = mg that probably appears in these 4-5 'learning experiences'. In fact it should read $$\vec F = m\vec g$$ and if you interpret that as ##F = mg\ \ ## which is a common notation for $$|\vec F| = m | \vec g |\ ,$$ you have basically thrown away the direction information with a 50% chance of error in 1D.

In a way, throwing away all direction info costs 50% in 1D, 75% in 2D and 87.5% in 3D .

As Peter says, you can make any choice, but once it's made you have to be consistent.

In many 1D exercises you naturally () assume up is positive. In such a case F = mg is true (it always is), but the interpretation ##\vec F_y = mg## is an error: ##\vec F_y## points upwards (which was chosen as positive 'direcetion) and ##\vec g_y## is down. Setting g = -9.81 m/s2 is asking for errors.

vanhees71
vanhees71
$$m \ddot{\vec{r}}=m \vec{g},$$
$$\vec{r}(t)=\frac{1}{2} \vec{g} t^2 + \vec{v}_0 t + \vec{r}_0.$$