A Question about a Quote from the Feynman lectures

[rudransh verma]

“incidentally, to a good approximation we have another law, which says that
the change in distance of a moving point is the velocity times the time interval, Deltas=vdeltat This statement is true only iF the Velocity is not changing during that time interval, and this condition is true only in the limit as Deltat goes to 0.
Physicists like to write it ds = vdt , because by dt they mean Deltat in circumstances in which it is very small; with this understanding, the expression is valid to a close approximation. If Deltat is too long, the velocity might change during the interval and the approximation would become less accurate. For a time dt, approaching zero, ds=vdt precisely. In this notation we can write (8.5) as
v=lim deltas/deltat = ds/dt. “

Now the deltas=vdeltat is true in itself. v is the average velocity. The sentence in bold is confusing me! When we take limit of delta t only then it changes to ds=vdt. The statement above is implying that deltas=vdeltat is wrong and ds=vdt as right.

 

[ergospherical]

What it means is that if $x = x(t)$ is a given function of time, then $\dfrac{x(t_0+h) - x(t_0)}{h}$ is not (generally) the velocity at $t_0$. As I'm sure you're already aware, that expression only goes over to $\dfrac{dx}{dt}(t_0) \equiv v(t_0)$ when you take $h \rightarrow 0$.

 

[rudransh verma]

What it means is that if $x = x(t)$ is a given function of time, then $\dfrac{x(t_0+h) - x(t_0)}{h}$ is not (generally) the velocity at $t_0$. As I'm sure you're already aware, that expression only goes over to $\dfrac{dx}{dt}(t_0) \equiv v(t_0)$ when you take $h \rightarrow 0$.

I don’t think that’s what it’s saying.In deltas=vdeltat , v is average velocity. The formula is true on its own. The formula of average velocity gives us average where it is preassumed that the velocities might be changing. That’s why it’s called average.

 

[ergospherical]

He's not talking about averages in that quote, although what you wrote is true! It's simply from the definition of the average of a function:\begin{align*}
\bar{v} = \dfrac{1}{\Delta t} \int_{t_0}^{t_0 + \Delta t} v(t) dt = \dfrac{1}{\Delta t} \left[ x(t) \right]_{t_0}^{t_0 + \Delta t} = \dfrac{x(t_0 + \Delta t) - x(t_0)}{\Delta t}
\end{align*}In this context $\Delta t$ is just some finite time interval.

 

[jbriggs444]

v is average velocity

Not in the context of the quote from Feynman.

In the context of Feynman's quote, $v$ is a function of time that may or may not vary. He seems to further assume that velocity varies smoothly over time. That is a reasonable assumption.

More formally, velocity is defined as the instantaneous rate of change of position with respect to time.

 

[rudransh verma]

In the context of Feynman's quote, v is a function of time that may or may not vary. He seems to further assume that velocity varies smoothly over time. That is a reasonable assumption.

You mean in deltas=vdeltat v is changing and so it’s not valid eqn. It will be if we take v as constant over deltat. And that is defined generally as average velocity in every textbook.

 

[jbriggs444]

You mean in deltas=vdeltat v is changing and so it’s not valid eqn.

I was writing only to clarify the meaning of v.

But yes, if $v$ does not denote a well defined value then it would be premature to include it in an equation.

I read $\Delta s = v \Delta t$ as $\Delta s = v(t_m)\Delta t$ for some unspecified time $t_m$ in the middle of the interval. Or maybe at its beginning, as has been suggested up-thread.

Note that, by the mean value theorem, there is a $t_m$ somewhere in the range from $t_i$ to $t_f$ (exclusive) that makes the equation true.

 

[rudransh verma]

Note that, by the mean value theorem, there is a tm somewhere in the range from ti to tf (exclusive) that makes the equation true.

That tm will give us the value of average velocity not instantaneous velocity which we would normally get by deltas/deltat ?

 

[hutchphd]

That tm will give us the value of average velocity which we would normally get by deltas/deltat !

Feynman explicitly defines v in eqn 8.5 immediately preceding:
$$v=\lim_{\Delta t\to0}\frac{\Delta s}{\Delta t}$$.Why do you think you are allowed to redefine it? It is defined in the limit. He even makes it a numbered equation.
There is nothing here.

 

[rudransh verma]

Feynman explicitly defines v in eqn 8.5 immediately preceding:
$$v=\lim_{\Delta t\to0}\frac{\Delta s}{\Delta t}$$.Why do you think you are allowed to redefine it? It is defined in the limit. He even makes it a numbered equation.
There is nothing here.

How is v in deltas=vdeltat instantaneous velocity when it is defined as average velocity. deltas/deltat=average velocity not instantaneous!

 

[hutchphd]

It is defined as the limit which is exactly the definition of the intantaneous velocity as pointed out by several others. That is what v is when Feynman writes it down.

 

[rudransh verma]

It is defined as the limit which is exactly the definition of the intantaneous velocity as pointed out by several others. That is what v is when Feynman writes it down.

So deltas=vdeltat is true if v(instantaneous) is same for every t in deltat which it never is in deltat. It may vary from point to point. It is constant in small time dt. So this statement is not a valid one. Correct one is ds=vdt when we take the limit.

 

[hutchphd]

What Feynman said is exactly correct. I honestly have no idea what you are saying.

 

[vanhees71]

How is v in deltas=vdeltat instantaneous velocity when it is defined as average velocity. deltas/deltat=average velocity not instantaneous!

Have you already learned what derivatives are? If not you should do so before reading any serious textbook on physics!

 

[rudransh verma]

Have you already learned what derivatives are?

Yes
Ok maybe now I get it. Generally it’s deltas=vaveragedeltat. But it can be deltas=vdeltat if velocity is not changing in interval deltat. And that is true when we take a small portion of deltat as dt by taking limit.

 

[vanhees71]

First of all the position vector, velocity, and acceleration are vectors. You define the position vector as a vector $\vec{x}$ pointing from some arbitrary fixed point in space, the origin of your reference frame, to the location of the point mass. To describe a motion of this point charge you consider $\vec{x}(t)$, i.e., the position vector as a function of time. In Newtonian mechanics space and time are just given (absolute space and absolute time) and invariable, i.e., they don't change by any physical process.

The velocity is then defined as the time derivative
$$\vec{v}(t)=\dot{\vec{x}}(t)=\frac{\mathrm{d}}{\mathrm{d} t} \vec{x}(t)$$
and the acceleration as the derivative of the velocity and thus the 2nd derivative of the position vector
$$\vec{a}(t)=\dot{\vec{v}}(t)=\ddot{\vec{x}}(t)=\frac{\mathrm{d}^2}{\mathrm{d} t^2} \vec{x}(t).$$
So velocity describes the momentaneous change of the position per unit time. Indeed, using the definition of a derivative you have
$$\vec{v}(t)=\lim_{\Delta t \rightarrow 0} \frac{\vec{x}(t+\Delta t)-\vec{x}(t)}{\Delta t}.$$
By the same principle it's clear that acceleration is the instantaneous change of velocity per unit time.

Using integrals you get the inverse of these operations:
$$\vec{x}(t)=\vec{x}(t_0) \int_{t_0}^t \mathrm{d} t' \vec{v}(t')$$
and analogously to get $\vec{v}(t)$ by an integral from $\vec{a}(t)$.

 

[rudransh verma]

First of all the position vector, velocity, and acceleration are vectors. You define the position vector as a vector x→ pointing from some arbitrary fixed point in space, the origin of your reference frame, to the location of the point mass. To describe a motion of this point charge you consider x→(t), i.e., the position vector as a function of time. In Newtonian mechanics space and time are just given (absolute space and absolute time) and invariable, i.e., they

Yeah ! But have I got it right?

 

[vanhees71]

I'm not sure, whether I understand you question right. It seems to be about the intuitive idea behind why the derivative of the position vector wrt. time is the "momentaneous velocity".

That works as follows: Consider you have your trajectory $\vec{x}(t)$ given (position vector as a function of time). Then the average velocity when you start at time $t$ and check where the point is at time $t+\Delta t$, where $\Delta t$ is some finite time increment than by definition the average velocity is
$$\langle \vec{v} \rangle_{\Delta t} =\frac{\vec{x}(t+\Delta t)-\vec{x}(t)}{\Delta t}.$$
That get's of course a pretty sparse information on your velocity if $\Delta t$ is large. So if you want to know the momentaneous velocity at time $t$, you have to make $\Delta t \rightarrow 0$, and then you have for any given time $t$ the momentaneous velocity vector as the time derivative of the position vector.

 

[jbriggs444]

Yes

A derivative is a limit of a ratio. It is not the ratio of two limits.

Ok maybe now I get it. Generally it’s deltas=vaveragedeltat. But it can be deltas=vdeltat if velocity is not changing in interval deltat. And that is true when we take a small portion of deltat as dt by taking limit.

Since we define $v_\text{avg}$ as $\frac{\Delta S}{\Delta t}$ it is true by definition that $\Delta S= v_\text{avg}\Delta t$. There is no physical content to the assertion.

 

[rudransh verma]

@jbriggs444 @vanhees71 I got it guys! It is simply saying when velocity is constant then average and instantaneous velocity will be equal.

 

[italicus]

When one begins studying derivatives in calculus, sooner or later meets three theorems : Rolle, Lagrange, Cauchy. In particular, I suggest the OP to have a look at the Lagrange theorem in this context