What does the term direction of motion actually mean? Is it a direction where a particle is moving or the direction of its velocity? For example, what is the direction of motion of a projectile in each case when it is moving upward and then downward? In each case, the direction of its acceleration is downward.

  • $\begingroup$ How can a projectile be moving upwards when its velocity is downwards? Maybe you are confused between displacement and it time average velocity $\frac{s}{t}$ and the velocity at a specific time $\frac{\delta s}{\delta t}$. $\endgroup$ – fibonatic Jan 24 '14 at 10:56

It means the direction defined by the velocity vector.

What is your definition of "a direction where a particle is moving" of not for the direction of its velocity vector?

Also, what you say about the projectile is false. The projectile moves along a parabolic path. Before it reaches its apex, its velocity vector has a vertical component pointing upward, and after it reaches its apex, its velocity vector has a vertical component pointing downward. The direction of its velocity is not downward the whole time. You may be thinking of its acceleration vector which does point downward the entire time.


The position of your projectile is a vector, $\vec{r}$, commonly given in component notation as $(x, y, z)$. The velocity, $\vec{v}$, of the particle is also a vector and is given by $d\vec{r}/dt$. The direction of motion is the direction the velocity vector points in.

To take your example of a projectile fired straight up, because only it's height, $z$, is changing the velocity vector will look like:

$$ \vec{v} = (0, 0, \frac{dz}{dt} ) $$

On the way up $dz/dt$ will be positive because the height is increasing, while on the way down $dz/dt$ will be negative because the height is decreasing, so comparing the velocity halfway up and halfway down (and ignoring air resistance):

$$ v_{up} = (0, 0, v_{1/2}) $$

$$ v_{down} = (0, 0, -v_{1/2}) $$

Where $v_{1/2}$ is the magnitude of the velocity halfway to the top of the trajectory.

So $v_{down} = -v_{up}$ - the two velocity vectors point in opposite directions not the same direction.


No that the velocity vector is always tangent to the path. So the direction of motion can be viewed as the direction of the velocity vector, as well as the direction of the path at that location. A path has direction because it has a start and a finish.

Mathematically if a path has position vector $\vec{r}(q)$ where $q$ is any independent coordinate such as distance, angle, time, then the direction of motion is defined by

$$\vec{e}(q) = \frac{\vec{r}'(q)}{|\vec{r}'(q)|} $$

where $\vec{r}'(q) = \frac{{\rm d}\vec{r}(q)}{{\rm d} q}$. The velocity vector is defined as

$$\vec{v}(q) = \vec{r}'(q) \;\dot{q}$$

So the direction of motion depends on the direction of the path $\vec{r}'(q)$ and the sign of $\dot{q}$ which indicates if you are moving forwards or backwards on the path.

Welcome to the world of differential geometry.


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