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?


2 Answers 2


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. enter image description here


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