I don't think the writer phrased it well. IMO, velocity is always positive regardless of the sign of acceleration (a, -a). The simplest verification is through the equation V = Vo + at. V can never be less than 0, which does not have a physical meaning. At least I am aware of it.

@r13, so what you are claiming is that velocity is always positive, and that the proof you are providing is $V=V0+at$. Posts like this make me wish I could downvote comments.

@r13 $V$ very much can be negative! A negative $V$ implies motion in the opposite direction to that of positive $V$. Besides, $V_0$ itself can positive or negative. Even if the initial velocity is positive, with a negative acceleration $a=-|a|$, it is clear from $V = V_0 - |a| t$ that the velocity goes negative after time $t = V_0/|a|$.

@r13 From the comments of DKNGuyen’s answer, it is clear you are mixing up velocity $V$ (a vector), and speed $|V|$ (a scalar). Even if the problem is 1D, velocity is very much still a vector, where, as mentioned before, the direction of which is dictated by sign (+1 and -1 are the unit vectors of 1D).

Numerically it can get to negative, but can one of you provide the physical explanation of Negative Velocity. Also, do we express distance from A to B is -3000', and time - he used -3s drove from A to B. Something just doesn't have a negative sense, though numerically it can.

@r13, throw a ball in the air, as it goes up then down, one of these needs to be negative. A negative velocity means backing up.

@Tiger When throwing the ball up, the "change" in velocity is negative, because the initial V >0 (non-negative) and Vf = 0 (non-negative). When it falls back, the "change" in velocity is positive, because Vo = 0 (non-negative) and Vf >0 (non-negative) before hitting the ground. So when the ball moves with a negative velocity?

@r13 I'm guessing that you are splitting up the problem into two different halves around the apex and solving them separately. If that's what you're doing you're solving two different problems and using two different coordinate systems in each problem. This won't work if you try to solve the problem as one continuous problem because then you need a consistent coordinate system that describes everything throughout.

What's happening is you're micro-managing the coordinate system based on what you know is already going to happen rather than letting the math figure it out on its own. This approach won't work for more complex problems where you don't know what is going to happen ahead of time or with multiple objects. In your description, you describe the velocity as always being positive, yet simultaneously you say the ball moves in one direction, then the opposite direction yet never changing sign.

@r13, the equation shows how velocity changes from positive to negative, just solve for velocity & graph it over time. Stop treating it like a non-scalar, it isn't.

@r13 I have already been clear in explaining the physical significance in the sign (+/-) of the velocity: opposing signs corresponds to opposing directions. You are indeed right to say that distance must be non-negative, but you are wrong to relate velocity to distance. Distance corresponds to speed, while displacement (or position) corresponds to velocity. It makes good sense for displacements to be either positive or negative, in the same way you might say something is, say, 5 metres forward (+5m) of some reference point or 5 metres backwards (-5m).

@r13 To be clear: if we represent displacement with $x$, then the definition of velocity is the rate of change of displacement (not the rate of change of distance! That would be speed!) or, mathematically, $V=\frac{\mathrm{d}x}{\mathrm{d}t}$. To deny the existence of a negative displacement would not be nothing short of sheer folly: it would be equivalent to saying that in graphs of $x$ and $y$, that $x < 0$ and $y < 0$ do not correspond to valid positions on a graph. Which I’m sure you will agree is madness!

So, consider $x$ to represent the height of a mass on a spring above a resting position. If $x$ is negative, this means the mass is below the resting point (if at this point you deny this much, then I implore you to better review your understanding on displacements and velocities!!). Consider the mass to oscillate, so its displacement as a function of time is $\sin{(\omega t)}$. Then, its velocity (according to the definition of velocity no less) is $\omega \cos{(\omega t)}$. This is a function whose sign oscillates between negative and positive, which corresponds to the mass having...

...downward and upward velocity respectively. In this context, negative velocity = downward velocity, positive velocity = upward velocity. To say velocity without direction is invalid, as that would just be speed. To encode the direction into the sign of the velocity for 1D problems is extremely convenient mathematically (you don’t have to break the problem up every time the object flips direction!). I have seen only a stubborn attempt to refuse this which I believe stems from a confused understanding of the definition of velocity.

Jone drove his car from A to B with the initial velocity -30 mph, and acceleration 2 m/s^2, find the final velocity after 10 s.

I’m going to assume you didn’t read my comments, but ok. In this context, I think it’s safe to assume that + means forward. So, Jone is driving backwards at 30m/s. Nothing odd there. He is accelerating forward (acceleration is a vector too!* acceleration is rate of change of velocity, and thus must also have direction! The positive value of acceleration implies acceleration forward) Roughly speaking, the positive acceleration “counteracts” the negative velocity, and so the car slows down. After the 10s, the car is now only going 10m/s backwards. Or -10m/s

So, quiz: you have a point in 3D space travelling in some unknown direction at a speed of 20m/s. (Now, the speed could never be -20m/s, because we know speeds cannot be negative!) So, you know the speed... are you able to tell me the velocity of this point?

I think we can agree that velocity is a physical vector quantity; "both magnitude and direction are needed to define it". A negative sign in front of the magnitude alone represents a mathematics operation with the negative operator (-) only.

@r13 “A negative sign in front of the magnitude alone represents a mathematics operation with the negative operator (-) only.” It’s almost like you are ignoring what I’m saying. So, in just the way you can express a velocity $\mathbf{V}$ in 3D as the speed $s$ (the magnitude) multiplied by a unit vector $\mathbf{e}$ (the direction), i.e. $\mathbf{V} = s\mathbf{e}$, in 1D you can represent the velocity $V$ as speed $s$ (magnitude) times direction +1 or -1, so $V = +s$ or $V = -s$ . It is up to you to define which direction that +ve represents, which is -ve.

So, no, the negative sign is not just a sign but significantly expresses direction! I don’t know what compels you to stubbornly insist in denying how sign cannot be used to represent velocity’s direction! I can’t help but feel like my words are wasted discussing something so fundamental in kinematics and so widely used in engineering.

The unit directional vector "e" is mathematically defined as "+1" or "-1" to represent the change of direction. So +V means (positive) velocity in the positive direction (+e), -V means (positive) velocity in the reverse direction (-e). I am quitting this great discussion, thanks for your effort.

What you’re calling $V$ here and restricting to only being positive (i.e. culling the direction) is not the velocity, but is the speed (Speed is the magnitude of velocity). It is $+V$ and $-V$ that are the velocities. Anyways, I wish you the best of luck.