# Is the direction of average velocity the same as that of average acceleration and that of displacement?

Average velocity is defined as: $\vec{\Delta v} = \frac{\vec{\Delta r}}{\Delta t}$, and average acceleration as $\vec{\Delta a} = \frac{\vec{\Delta v}}{\Delta t}$.

It is apparant from these definitions that average velocity, average acceleration and displacement all have the same direction, but I have got stuck in this:

Assume that we are studying the motion of a particle in one dimension:
Let $\Delta t = 1s$, $\vec{v_i} = -4 \hat{\textbf{i}} m/s$ and $\vec{v_f} = -2 \hat{\textbf{i}} m/s$.
Hence, $\vec{\Delta v} = 2 \hat{\textbf{i}} m/s$ and $\vec{\Delta a} = 2 \hat{\textbf{i}}$, so both of average-velocity's and average-acceleration's direction are toward the positive $x$-axis, but surely the displacement will be toward the negative $x$-axis since both velocity values are toward the negative $x$-axis. Is what I have guessed right?

You are on the right track, but $$\Delta\vec{v}$$ is not average velocity. It's just the change in velocity:

$$\Delta\vec{v} = \vec{v_f} - \vec{v_i} \neq \frac{\Delta \vec{r}}{\Delta t}.$$

For your example, find $$\Delta\vec{r}$$ from a kinematic equation:

$$\Delta\vec{r} = \vec{v_i}\Delta t + \frac{1}{2}\vec{a}\Delta t^2 .$$

With your numbers, $$\Delta \vec{r} = -3\hat{i}$$ m, and $$\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} = -3\hat{i} \frac{\textrm{m}}{\textrm{s}}$$.

The displacement can be positive while $$\vec v$$ and $$\vec a$$ are negative. Imagine a car going from $$x=10$$ to a streetlight at $$x=0$$. At every $$t$$ the displacement is positive since the car is always on the $$x>0$$ side. However surely $$\vec v$$ is negative since the displacement is from $$x=10$$ towards $$x=0$$, i.e. while $$\vec r$$ is positive, $$\Delta \vec r$$ is negative.

Moreover, in the case of 1d motion, it is entirely possible to have constant negative acceleration yet positive or negative velocities. This is best captured by the equation $$v(t)=v_0-a t\, ,\qquad a>0$$ so that, if - say - the initial velocity $$v_0= 10m/s$$ while $$a=5m/s^2$$, we find that between $$0\le t\le 2$$, the velocity is positive while for $$t>2$$ the velocity is negative. Of course, because $$v(t)$$ changes sign during the motion, there will be time intervals where $$\Delta \vec r>0$$ and times when $$\Delta \vec r<0$$ since the equation for position as a function of time is (in 1d) $$x(t)=x_0+v_0t-\frac{1}{2}at^2= x_0+10 t -\frac{5}{2}t^2\, ,$$ which will be positive (for $$x_0=0$$) for small $$t$$ but negative for larger $$t$$, i.e. the object will pass through $$x=0$$ at $$t=0$$ and $$t=4$$.

Actually one should not think that both $$\Delta \vec v$$ and $$\Delta \vec a$$ need to be in the same direction or even co-linear. Think of circular motion: then clearly the average velocity is (for sufficiently short times) nearly tangent to the circle but the average acceleration will be directly basically towards the centre of the circle.

A simple example is provided assuming the Moon is in circular motion about the Earth. Clearly the force $$\vec F_{ME}$$ will be co-linear with the acceleration but obviously here this acceleration is not in the same direction as the velocity.