High School Understanding Galileo's Theory of Acceleration

[Huzaifa]

I am not able to understand the following paragraph from my Physics textbook;

"The velocity of an object, in general, changes during its course of motion. Should it be described as the rate of change in velocity with distance or with time? This was a problem even in Galileo's time. It was first thought that this change could be described by the rate of change of velocity with distance. But, through his studies of motion of freely falling objects and motion of objects on an inclined plane, Galileo concluded that the rate of change of velocity with time is a constant of motion for all objects in free fall. On the other hand, the change in velocity with distance is not constant - it decreases with the increasing distance of fall. This led to the concept of acceleration as the rate of change of velocity with time."

 

[.Scott]

This discussion addresses a body in free fall - and it specifically addresses the effects of gravity.
For the kind of small-scale experiments that Galileo was using (non-orbital), the position of an object in free-fall can be described with a constant acceleration value $a$. The formula for the height ($y$) is ($y = at^2/2$) - thus it is a function of time ($t$).

The vertical velocity ($v$) as a function of time ($t$), the starting velocity ($v_0$), and acceleration due to gravity ($a$) is quite simply: $v=v_0+(at)$.

But before Galileo deduced these formulae, it was possible that this change in velocity would turn out to be easier to express as a function of distance. Careful experimentation nailed the formula - and it turned out to be a function of $t$.

 

[hutchphd]

The vertical velocity (v) as a function of time (t) and the starting velocity (v0) and acceleration due to gravity (a) is quote simple: v=v0+(at2/2).

Please correct this. Also I don't really understand your explanation

 

[.Scott]

Please correct this. Also I don't really understand your explanation

First: I notice that when you quoted my post, you didn't pick up the LaTex correctly - or at least it is not showing correctly in your post as I got it. Hopefully, that is not a cause for confusion.

I fixed the formula (oops) and a few typos (ex, "Care" is now "Careful").

The original explanation provided to @Huzaifa (in post #1) looked good to me. I restated parts of it with a bit more detail - hoping that @Huzaifa would catch on. It's always difficult to try to guess why someone doesn't catch something. In this case, I suspect that the problem is missing pieces of context. But we will only know when @Huzaifa says "got it".

 

[sophiecentaur]

Careful experimentation nailed the formula - and it turned out to be a function of t.

I don't think that's necessarily a universal rule, though. Galileo was studying a particularly simple form of accelerated motion. Isn't is just a matter of the process of the Maths involved? If you take a compound pendulum (or an arrangement of a mass on multiple springs) the force (=acceleration) is a direct function of position and time can be a bit secondary - perhaps not really relevant. There would be alternative ways of presenting a description of the motion and, often, time is a good way but the path is often more convenient form of description. The destination of a shell's trajectory is what counts and time is only a step on the way to the answer.

But this is all a matter of opinion and can lead us to Top Trumps if we're not careful.

 

[.Scott]

I don't think that's necessarily a universal rule, though. Galileo was studying a particularly simple form of accelerated motion.

According to the OP:

... through his studies of motion of freely falling objects and motion of objects on an inclined plane, Galileo concluded ...

So I was answering within the context of acceleration due to gravity.
Acceleration due to friction, rocket engines, etc would each have their own equations.

 

[Herman Trivilino]

I am not able to understand the following paragraph from my Physics textbook;

It's a discussion of which of two definitions is more useful. If you divide a change in velocity by the distance an object travels during that change in velocity you get a quantity that is less useful.

 

[Delta2]

However in some problems, the differential equation for velocity/speed is such that we can find a close form for the velocity as function of distance $v(x)$ but can't find a close form for the velocity as function of time $v(t)$.

It is on these problems that we use the chain rule trick to get rid of $\frac{dv}{dt}$ as $$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$$

 

[vanhees71]

Usually this form of the equations of motion occurs when you have an effectively one-dimensional problem and energy conservation holds. Then you have something like (here written for a usual cartesian component $x$ and the corresponding velocity $v=\dot{x}$)
$$\frac{m}{2} v^2+V(x)=E.$$
Then you can solve for $v(x)$
$$v(x)=\pm \sqrt{2m [E-V(x)]}.$$
From this you get a solution in implicit form since $\dot{x}=v(x)$ can be integrated by "separation",
$$t-t_0=\pm \int_{x_0}^x \frac{1}{\sqrt{2m[E-V(x)]}}.$$
You have to carefully deal with the signs of the square root to get the correct solution!