# What is the difference between these two ways to calculate average velocity?

Average velocity:

$$v_{\rm avg,1}=\frac{v_{\rm final}+v_{\rm initial}}{2}$$

and average velocity:

$$v_{\rm avg,2} =\frac{\rm total\;displacement}{\rm time \;taken}=\frac{\Delta x}{\Delta t}$$

What is the difference between them and when do we use them?

• Related: physics.stackexchange.com/q/55809/2451 , physics.stackexchange.com/q/44685/2451 and links therein. – Qmechanic Aug 14 '16 at 14:05
• First one doesn't work when the rate-of change of velocity (with time) is not uniform. Say a train initially go for 0.01 m/s for 1s . then 1 m/s for 10 second, and then 5m/s for 10 days. You can't say the average velocity by the fist way. The ans. should be slightly less than 5 m/s. You can determine it from 2nd formula. – Always Confused Aug 14 '16 at 15:13
• danny, remember you may upvote the answers that have been useful and that you should accept the most helpful one (and you earn 2 rep points) clicking on the check sign beneath votes. ( if you don't know how,click on help abobe, near search – user104372 Oct 7 '16 at 8:22

Your first way of calculating an average velocity is inaccurate and really should be avoided.

Only the second method is accurate. This is a consequence of the underlying calculus of kinematics.

When a object travels (e.g. but not necessarily on a straight line) its velocity is not required to be constant. In fact for the general case we assume $v$ is a function of time, mathematically noted as:

$$\Large{v(t)}$$

Physically the velocity is the first derivative of position ($x$) to time ($t$):

$$\Large{v(t)=\frac{dx}{dt}}$$

To find the displacement $\Delta x$ during an interval of time $\Delta t=t_2-t_1$ then $\Delta x$ is calculated by integration:

$$\Large{\Delta x=\int_{t_1}^{t_2}v(t)dt}$$ This also means that the average velocity $\bar{v}$ can be calculated from: $$\Large{\bar{v}=\frac{\Delta x}{\Delta t}}$$

This is true regardless of how $v(t)$ evolves over the time interval $\Delta t$.

Taking the "average" by averaging two velocity readings however is meaningless.

In response to OP's comment about constant acceleration:

If acceleration is constant the velocity is given by:

$$v=v_0+at$$

Where $v_0$ is the velocity at $t=0$.

After a time interval $\Delta t$ the velocity has become:

$$v_1=v_0+a\Delta t$$

The displacement would be:

$$\Delta x=\int_0^{\Delta t}(v_0+at)dt=v_0\Delta t+\frac12 a(\Delta t)^2$$

The average velocity $\bar{v}$ is:

$$\bar{v}=\frac{\Delta x}{\Delta t}=v_0+\frac12a\Delta t$$

Using the first method:

$$\bar{v}=\frac{v_0+v_0+a\Delta t}{2}=v_0+\frac12a\Delta t$$

So that in the case of constant acceleration we obtain the same result. Note that this is the only case where both give the same result.

• Can't we use the first when an object is undergoing projectile motion – danny Aug 14 '16 at 14:14
• Can't we use it when velocity is constant – danny Aug 14 '16 at 14:15
• Hi Danny. If velocity is constant then what's the point in taking an average? $(v_1+v_1)/2=v_1$! ;-) – Gert Aug 14 '16 at 14:17
• Sorry i meant to say if acceleration is constant – danny Aug 14 '16 at 14:20
• I'll answer that in a edit to the post. – Gert Aug 14 '16 at 14:21

Taking the average of the initial velocity and final velocity is not necessarily, you are assuming a linear change in the velocity which is not the general situation. So only the second formule specifies the average velocity.

The correct equation for the average velocity is the second equation of the both equations you gave. The first one is correct only under given condition that the acceleration of the body is constant. Second equation even holds for variable acceleration.

The average velocity of a particle during some elapsed time $\Delta t$ is, in words, the constant velocity that gives the same displacement in the same elapsed time.

Mathematically, the average velocity is given by

$$\mathbf{v}_{avg} = \frac{\Delta \mathbf{r}}{\Delta t}$$

where $\Delta \mathbf{r} = \mathbf{r}_f - \mathbf{r}_i$ is the displacement vector and $\Delta t = t_f - t_i$ is the elapsed time during which the displacement took place.

For example, consider the case that a particle moves with constant velocity $1 \mathrm{\frac{m}{s}} \hat{\mathbf{x}}$ for 4 seconds and then with constant velocity $1 \mathrm{\frac{m}{s}} \hat{\mathbf{y}}$ for 3 seconds.

The displacement vector for the 7 seconds of motion is, by inspection,

$$\Delta \mathbf{r} = (4\hat{\mathbf{x}} + 3 \hat{\mathbf{y}})\;\mathrm{m}$$

and so, the average velocity during the 7 seconds is

$$\mathbf{v}_{avg} = (\frac{4}{7}\hat{\mathbf{x}} + \frac{3}{7} \hat{\mathbf{y}})\;\mathrm{\frac{m}{s}}$$

Clearly, if another particle had this constant velocity and started at the same initial point at the same time as the first particle, the two would reach the same final point at the same time.

On the other hand, the quantity

$$\frac{\mathbf{v}_f + \mathbf{v}_i}{2}$$

is an average of two velocities, which is not particularly useful or meaningful, not an average velocity which has a clear and useful meaning.

There are two special cases:

(1) In the case that the particle spends half of the elapsed time at a constant velocity $\mathbf{v}_1$ and spends the other half of the elapsed time at a constant velocity $\mathbf{v}_2$, then the average velocity is just the average of the two velocities.

(2) In the case that the particle has constant acceleration, the velocity increases linearly with time and so the displacement per unit time (and working in 1-D)

$$\Delta r = v_i + \frac{(v_f - v_i)}{2} = \frac{v_i + v_f}{2}$$

and thus, the average velocity is just the average of the initial and final velocities.