Calculate Average Velocity: vav = (d2 - d1)/(t2 - t1)

• AnthonyC
In summary: Second, although the average velocity does occur at the midpoint of the time interval, it is not always the instantaneous velocity. Third, the formulas given to you are not always the same, and in this particular case, the one given is for a particle moving under constant acceleration.
AnthonyC
I am doing this question, and have collected the times and distances traveled at those various times.

I am supposed to calculate the average velocity for each distance now.

The book say: vav = vinst = d/t ; at t’ = t/2 (half time)

But I checked online and found: vav = (d2 – d1)/(t2 – t1)

Earlier in the book, it indicated that it was: vav = d/t.

Which is it?!

This doesn't make sense to me!

It certainly does not make any sense! vav = (d2 – d1)/(t2 – t1) makes the more sense. And notice that the book's version's, vav = d/t is pretty much the same thing, except it's for the special case where d1 = 0 and t1 = 0. But vav = vinst = d/t ; at t’ = t/2 (half time) is completely off track. It must have been written in the context of a very specific case.

I know, the question says:

"3. Calculate the average velocity for each distance

NOTE: The average velocity occurs at the mid-time point of the time interval over which the measurement was taken (see Average and Instantaneous Speed Module I - Uniform Accelerated Motion)

vav = vinst = d/t ; at t’ = t/2 (half time)"

I don't know what I should do, I suppose use the formula they provided as I at least have a legitimate argument if they mark it wrong, correct?

All the distances measured started from 0, so
0 to 1 = .004,
0 to 2 = .015,
0 to 3 = .032,
etc,

And all the time intervals were 0.05 s apart,
0.050 s,
0.100 s,
0.150 s,
etc.

Ok, I just realized that the little formula is true for constant acceleration. I will show you why.

But I don't see how it is helpful to you here. Just use

$$v_{av} = \frac{x_2-x_1}{t_2-t_1}$$

P.S. Are you sure that the formula doesn't read $v_{av} = \frac{dd}{dt}$ at t' = t/2 instead?

Last edited:
This is the exact equation they say to use:

vav = vinst = d/t ; at t’ = t/2 (half time).

Very confusing!

Later on, I am supposed to find the acceleration for each two consecutive time intervals.

Using: a_1 = (v_2 – v_1)/(t_2’ –t_1’)

So using that formula, for the first row of information (t' = .025s, v = .080 m/s) should I make v_1 = 0m/s and t_1 = 0s. Since this would be at rest?

The formula giving the position of a particle under constant acceleration as a function of time is

$$x(t) = x_0 + v_0t + \frac{a}{2}t^2$$

In the special case where $x_0 = v_0 = 0$, it becomes

$$x(t) = \frac{a}{2}t^2$$

$$v_{av} = \frac{x_2-0}{t_2-0} = \frac{x(t_2)}{t_2} = \frac{\frac{a}{2}t_2^2}{t_2} = \frac{at_2}{2}$$

On the other hand, the instantaneous velocity at time $t' = t_2/2$ is the derivative

$$\left[ \frac{dx}{dt}\right]_{t_2/2} = \left[at\right]_{t_2/2} = \frac{at_2}{2}$$

Last edited:
Wow, that's complex, lol.

Not sure if I know how to manipulate that for my situation.

I have the acceleration going from rest to .08 m/s in .025 s (t') being 3.20 m/s/s. Is that correct?

Anyway,you're actually computing the velocity's average:

$$\langle \vec{v}\rangle_{\left[t_{1},t_{2}\right]}=:\frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}} \vec{v}(t) \ dt$$

Daniel.

AnthonyC said:
This is the exact equation they say to use:

vav = vinst = d/t ; at t’ = t/2 (half time).

Very confusing!

Well.. I don't see how this is true. Because by definition, d = x(t). So

$$\left[\frac{d}{t}\right]_{t'=t_2/2} = \frac{x(t_2/2)}{t_2/2}=\frac{\frac{a}{2}(t_2/2)^2}{t_2/2} = \frac{at_2}{4}$$

AnthonyC said:
Later on, I am supposed to find the acceleration for each two consecutive time intervals.

Using: a_1 = (v_2 – v_1)/(t_2’ –t_1’)

So using that formula, for the first row of information (t' = .025s, v = .080 m/s) should I make v_1 = 0m/s and t_1 = 0s. Since this would be at rest?

Don't mix up the formulae. The primes are to be used when evaluating the derivative of position in order to find the average velocity (like in my proof above). If you want the average acceleration (which is also the instantaneous acceleration in the case of constant acceleration), just use the formula you wrote, except use the regular t_2 and t_1 instead of t_2' and t_1'.

Last edited:
AnthonyC said:
I know, the question says:

"3. Calculate the average velocity for each distance

NOTE: The average velocity occurs at the mid-time point of the time interval over which the measurement was taken (see Average and Instantaneous Speed Module I - Uniform Accelerated Motion)

vav = vinst = d/t ; at t’ = t/2 (half time)"

I don't know what I should do, I suppose use the formula they provided as I at least have a legitimate argument if they mark it wrong, correct?

All the distances measured started from 0, so
0 to 1 = .004,
0 to 2 = .015,
0 to 3 = .032,
etc,

And all the time intervals were 0.05 s apart,
0.050 s,
0.100 s,
0.150 s,
etc.

I think I understand the confusion surrounding this exercise, so I'm going back to this post as a starting point. First, although primes are often used to indicate derivatives, that is not what is going on here. The primes are just being used to distinguish times at which velocities are being calculated from times at which positions are given.

The confusing statement

vav = vinst = d/t ; at t’ = t/2 (half time)"

is valid under certain conditions that have only been vaguely stated. The truth is that if you know the position at two times you can always calculate the average velocity as the distance between the two positions divided by the time interval between them, as others have noted. There is also a theorem that says the average velocity for that interval will be equal to the instantaneous velocity at some moment during that interval. In a constant acceleration problem, the average velocity for any interval is the same as the instantaneous velocity at the middle of the time interval. The point of this exercise is that when you calculate an average velocity for each of the intervals of 0.05 second duration, the velocity you are calculating is both the average velocity for the interval, and the instantaneous velocity at the middle of the time interval.

You have a set of data giving postion at equally spaced times. You probably have a graph of that data. You can calculate the average velocity between each pair of adjacent points. The average velocity is not the instantaneous velocity at the beginning of the interval, nor at the end of the interval. It is, however, the instantaneous velocity at the point in time in the middle of that interval. If you want to graph the velocity, the time points you should use are the times in the middle of each interval (.025, .075, etc) The t' = t/2 is true if the reference time at the beginning of every interval is taken to be zero.

Once you graph your velocities, or tabulate them using the mid-point times, you can then calculate the accelertion for each interval. If you have a graph, you can see that the change in velocity is the same for each interval. If you don't have the graph, the calculation will show you that the change in velocity is the same for each interval. The accelertion is the slope of the velocity graph, or the average over the nearly constant values (only "nearly" because of round off or measurement errors) you get from the calculation. I expect the ultimate aim of the exercise is for you to come up with the acceleration of the object from the data you have of position versus time.

What is average velocity?

Average velocity is a measurement of the overall rate of change in an object's position over a certain period of time. It takes into account both the distance traveled and the time it took to travel that distance.

How do you calculate average velocity?

To calculate average velocity, you need to know the distance traveled (d2 - d1) and the time it took to travel that distance (t2 - t1). You then divide the distance by the time to get the average velocity (vav = (d2 - d1)/(t2 - t1)).

What are the units for average velocity?

The units for average velocity are distance over time, commonly written as meters per second (m/s) or kilometers per hour (km/h). However, other units such as feet per second (ft/s) or miles per hour (mph) can also be used depending on the distance and time measurements.

Can average velocity be negative?

Yes, average velocity can be negative. This means that the object is moving in the opposite direction of the positive axis or reference point. For example, if an object is moving to the left, its average velocity will be negative if the positive axis or reference point is to the right.

How is average velocity different from instantaneous velocity?

Average velocity is a measure of the overall rate of change in an object's position over a certain period of time, while instantaneous velocity is the rate of change in an object's position at a specific moment in time. Average velocity takes into account the entire journey, while instantaneous velocity only looks at a single point along the journey.

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