2
$\begingroup$

A simple function with vertical tangents was made with semicircles

Are vertical tangent lines possible in position-time diagrams of motions?

Since a vertical tangent in a position-time diagram would correspond to "infinite speed", I guess the answer is no. And if so, then the graph of the above function, made by connecting semicircles, could never be a correct position-time graph of a motion. But, how to show that this graph is impossible, let's say in Newtonian mechanics, i.e. that there is no possible situation that would result in the position being a function of time in such a way?

$\endgroup$
  • $\begingroup$ As you correctly state, vertical tangents are impossible because they would imply infinite velocity. So, what you want to prove is that the velocity becomes infinite when the curve intersects the $t$ axis in this particular case? $\endgroup$ – valerio Oct 9 '16 at 9:57
  • $\begingroup$ I would like to know how to show that "infinite speed" is impossible in Newtonian mechanics. I gave the graph just as an example, but showing that this particular one is impossible, in the sense that there is no situation, a system of particles with given initial conditions, that could result in this, would be nice too. I don't know if physics.stackexchange.com/questions/13184/… answers my question. Sorry, I don't really know if what I asked made sense, because I am confused. $\endgroup$ – Viktor Kaspervich Oct 9 '16 at 10:09
2
$\begingroup$

In Newtonian mechanics, including various extensions, like Lagrangian mechanics and Hamiltonian mechanics, the fundamental law determining the equation of motion of a system can always be stated into the form of a system of differential equation of second order in normal form. Namely, $$\frac{d^2X_i}{dt^2}= F_i\left(t,X_1(t), \ldots, X_N(t), \frac{dX_1}{dt}(t), \ldots, \frac{dX_N}{dt}(t)\right)\:, \quad i=1,\ldots, N\tag{1}$$ where $X_i$ are the coordinates also in a generalized sense of the points of the system. Provided the functions $F_i$ are sufficiently regular, the system (1) admits a unique solution in a set including $t_0$ when initial conditions $$\left(X_1(t_0), \ldots, X_N(t_0), \frac{dX_1}{dt}(t_0), \ldots, \frac{dX_N}{dt}(t_0)\right) = \left(X_{10}, \ldots, X_{N0}, \dot{X}_{10}, \ldots, \dot{X}_{N0}\right)$$ are given.

Since the functions $X_i=X_i(t)$ must be differentiable up to the second order, the first derivatives must be defined at every instant $t$ where the motion is defined. As a consequence none of $\frac{dX_i}{dt}(t)$ can attain an infinite value.

The graphical representation you discuss is referred to the case $N=1$. The instants of time where the tangent is vertical are associated to infinite velocities and this is not permitted.

Obviously, limit, unphysical strictly speaking, cases can always be produced assuming the functions $F_i$ (essentially the forces acting to the system) affected by some pathology.

$\endgroup$
0
$\begingroup$

No. A vertical section in a position-time graph would mean that the object (assumed to be a point particle) occupies more than one position at the same time. This is not possible. It can occupy one position at many different times, but it cannot occupy more than one position at one instant in time.

If the object is not a point particle, the graph must represent the motion of one point on the object - eg the centre of mass. An extended object might also rotate, but the graph cannot show this.

Of course the graph can get as close to vertical as you wish, depending on what scale you use for plotting. But it can never be exactly vertical.

Showing that "infinite speed is impossible in Newtonian Mechanics" has nothing to do with the distance-time graph. It has nothing to do with physics either, because "infinite" is not defined in physics. Any speed which you can write down, no matter how large, can be reached in Newtonian Mechanics, given enough acceleration and time.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.