# Conceptual problem in average/instantaneous velocity

In summary, the average velocity over a finite time interval is simply the average of the instantaneous velocities.

## Homework Statement

I am having a bit of confusion regarding this:
When you write dr/dt (rate of change of position vector with time, sorry I don't know how to use LATEX), do you mean instantaneous or average velocity?
As we are considering a very small time interval, I guess it is instantaneous. Now, can we define average velocity in the same way? (Δr/Δt which spans over a bigger time interval)
However, I also saw a few problems on calculating average velocity over a time interval in projectile motion, and they are calculating it by differentiating (ucosθt i + usinθt-0.5gt2 j), which is the position vector at an instant. Now, if instantaneous velocity is dr/dt, and not average velocity, then how can one proceed like this?

## The Attempt at a Solution

pranjal verma
Did they integrate it over time though ?

## Homework Statement

I am having a bit of confusion regarding this:
When you write dr/dt (rate of change of position vector with time, sorry I don't know how to use LATEX), do you mean instantaneous or average velocity?
As we are considering a very small time interval, I guess it is instantaneous. Now, can we define average velocity in the same way? (Δr/Δt which spans over a bigger time interval)
However, I also saw a few problems on calculating average velocity over a time interval in projectile motion, and they are calculating it by differentiating (ucosθt i + usinθt-0.5gt2 j), which is the position vector at an instant. Now, if instantaneous velocity is dr/dt, and not average velocity, then how can one proceed like this?

## The Attempt at a Solution

The average velocity over some finite time interval ##\Delta t## is simply ##\frac{\vec{r}(t + \Delta t) - \vec{r}(t)}{\Delta t}##.

##\frac{d \vec{r}}{dt}## is physically an instantaneous velocity; and, mathematically the time derivative of the postion function ##\vec{r}(t)##, which is the limit of the average velocity as ##\Delta t \rightarrow 0##:

##\frac{d \vec{r}}{dt} = \lim_{\Delta t \rightarrow 0} \frac{\vec{r}(t + \Delta t) - \vec{r}(t)}{\Delta t}##

If position (r) is some function of time then dr/dt gives you an equation for the velocity (v) as a function of time. To obtain the actual velocity at a point in time you have to substitute that value of time into the equation. So yes it's the instantaneous velocity.

Averaging typically involves integrating ("adding up") the velocity between two times (which could be, but doesn't have to be, the start and finish times) and then dividing by the time interval.

With problems like this you need to be careful of the time period over which you integrate/average. Eg they may ask you for the average velocity over the last 50 miles of a 100 mile trip which is not necessarily the same as the average velocity over the last half of the trip in terms of time.

Some definitions (amplifying the earlier responses):
• $\vec v(t)=\frac{d}{dt}\vec r(t)=\displaystyle\frac{d \vec r}{dt}$ is the [instantaneous-]velocity, a function of $t$.
• On a position-vs-time graph in 1D, at a given time $t$, it's the slope of the tangent line at time $t$.
• $\vec v_{av}(t_A,t_B)=\displaystyle\frac{\int_{t_A}^{t_B} \vec v\ dt }{\int_{t_A}^{t_B} \ dt } =\frac{\vec{r}_B-\vec {r}_A}{t_B-t_A}=\frac{\Delta \vec r}{\Delta t}$ is the "average-velocity", which is shorthand for the more complete:
[time-weighted-]average of instantaneous-velocities [over a time interval].
• In my opinion, many misconceptions arise because
• "average-velocity" is given more emphasis and introduced first before "instantaneous-velocity" and
• the time weighting feature of "average velocity" is not given.
• On a position-vs-time graph in 1D, it's the slope of the [secant] segment joining $(x_A,t_A)$ to $(x_B,t_B)$.
Instead of making the trip from $(x_A,t_A)$ to $(x_B,t_B)$
with the generally varying velocity for $\vec r(t)$,
use a constant velocity to start and arrive at the same places at the same times--that's the average-velocity.
• These coincide when $\vec r(t)$ has constant velocity.
• In the piecewise-constant case, this is $\vec v_{av}(t_A,t_B)=\displaystyle\frac{\int_{t_A}^{t_B} \vec v\ dt }{\int_{t_A}^{t_B} \ dt } = \frac{ \vec v_{AC}\Delta t_{AC}+\vec v_{CB}\Delta t_{CB} }{\Delta t_{AC}+\Delta t_{CB}} =\frac{ \Delta\vec r_{AC}+\Delta\vec r_{CB} }{\Delta t_{AC}+\Delta t_{CB}} =\frac{ \Delta\vec r_{AB} }{\Delta t_{AB}}$
Try it for projectile motion:
using the definition of average-velocity
with the instantaneous-velocity-formulas for projectile motion,
compute the average-velocity over an interval.

## 1. What is the difference between average velocity and instantaneous velocity?

Average velocity is the total displacement divided by the total time, while instantaneous velocity is the velocity at a specific moment in time.

## 2. How do you calculate average velocity?

To calculate average velocity, divide the total displacement by the total time taken to cover that distance. The formula is: average velocity = (final position - initial position) / (final time - initial time).

## 3. What is a conceptual problem in average/instantaneous velocity?

A conceptual problem in average/instantaneous velocity refers to a problem where the concept of velocity is being tested, rather than the mathematical calculation. It requires an understanding of the difference between average and instantaneous velocity and how they relate to each other.

## 4. How does changing the time interval affect average and instantaneous velocity?

Changing the time interval can affect the average and instantaneous velocity in different ways. For average velocity, a longer time interval will result in a lower average velocity, while a shorter time interval will result in a higher average velocity. For instantaneous velocity, a shorter time interval will result in a more accurate measurement of the velocity at a specific moment in time.

## 5. Can average velocity ever be equal to instantaneous velocity?

Yes, if the object is moving at a constant velocity, then the average velocity will be equal to the instantaneous velocity at any given moment. This is because the velocity does not change over time, so the average velocity and instantaneous velocity will be the same.

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