# Conceptual problem in average/instantaneous velocity

1. 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?

2. Homework Equations

3. The Attempt at a Solution

• pranjal verma
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#### TachyonLord

Did they integrate it over time though ?

#### PeroK

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2018 Award
1. 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?

2. Homework Equations

3. 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 $\Delta t \rightarrow 0$:

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

#### CWatters

Homework Helper
Gold Member
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.

#### robphy

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Gold Member
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.