The change in velocity can be calculated by vector subtraction. ($d\vec{v} = \vec{v_f} - \vec{v_i}$).
Divide by the time between the two velocities to generate an acceleration. The direction of the acceleration will be the same as the direction of the difference vector. The magnitude of the acceleration will be the same as the magnitude of the difference vector divided by the time.
Example, at time $t=0$ seconds the velocity is in the $x$ direction, and at time $t=2$ seconds the velocity is in the y direction. At both times the velocity is 1 m/s:
$$\vec{v(0)} = (1,0,0) ~\mathrm{m/s}$$
$$\vec{v(2)} = (0,1,0) ~\mathrm{m/s}$$
$$d\vec{v} = \vec{v(2)} - \vec{v(0)}$$
$$d\vec{v} = (-1,1,0) ~\mathrm{m/s}$$
$$dt = 2 - 0 = 2 ~\mathrm{s}$$
$$\vec{a} = \frac{d\vec{v}}{dt} = (-\frac12,\frac12,0) ~\mathrm{m/s^2}$$
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.
Best Answer
The equation is correctly telling you that $s = 0$.
Suppose you start at $x = 0$ and travelling in the positive $x$ direction with some velocity $v_0$. By the time you've accelerated to a halt (i.e. $v = 0$) you'll be at some positive $x$ value. You continue the acceleration to start moving in the opposite direction, and of course your position on the $x$ axis is now decreasing. By the time your speed is $-v_0$ you'll be back where you started i.e. $s = 0$. The equation is not telling you the total distance travelled, it's telling you the net distance travelled, and because you end up back where you started then net distance is zero.
To get the total distance travelled is easy. Just solve your equation for $u = v_0$ and $v = 0$, to find out how far you move before stopping, then double that distance if you want the total distance moved.