Can motion graphs have sharp edges?

Question

I'm pretty convinced but I need to make sure I'm right about this.

By a sharp edge, I mean a point in the graph where the curve is not differentiable.

in motion graphs, (x,y,z) coordinates depend on t polynomially.

All quantities in motion depend on coordinates and t.

Since operations on differentiable functions result in a differentiable function, motion graphs cannot have sharp edges (points where the curve cannot be differentiated)

Is this right?

Possible duplicates: https://physics.stackexchange.com/q/151399/2451 , https://physics.stackexchange.com/q/1324/2451 and links therein. — Qmechanic, May 03 '20 at 22:43

score 4 · Accepted Answer · answered May 04 '20 at 02:27

There are two answers to this question, depending on whether you are an engineer or a physicist, as follows.

In an idealized (physics) world where a ball of infinite stiffness strikes a surface of infinite stiffness and mass, the curve will be discontinuous, as you suspect.

In a messy (engineering) world, bouncing objects are not infinitely stiff, and if you smack two very stiff objects into one another, the system will "find" a compliance somewhere which it can compress so the impact force will not become infinite during the collision. This means that the curve will not be discontinuous and would in fact be differentiable.

score 3 · Answer 2 · answered May 03 '20 at 18:34

3

in motion graphs, (x,y,z) coordinates depend on t polynomially.

That is only a special case (but it may be the only thing you have seen, if you have just begun studying projectile motion).

The motion of rigid bodies when they collide is a standard situation where the "motion graph" is not differentiable.

answered May 03 '20 at 18:34

alephzero

10,129

I wonder, isn't it theoretically impossible to have rigid bodies? – user12986714 May 04 '20 at 04:02

score 2 · Answer 3 · answered May 03 '20 at 18:33

Consider a case in which a charged point sized particle is moving towards another due to electrostatic fore. Its velocity-time graph will have a non-differentiable point at the point where the the two charges meet-the acceleration will be undefined. So, motion graphs can have points where curve cannot be differentiated. (it is not a sharp edge, but still, you have specified non-differentiable points)

You can also imagine a case when the equations describing motion suddenly and abruptly change. A classic example (if you have done a course on animation, you'd know better), is a ball bouncing. The distance time curve has a sharp point whenever the ball hits the ground.

Can motion graphs have sharp edges?

3 Answers3