# How can this object accelerate if its position/time graph has no non-zero second derivative at any point?

This has puzzled me for a long time.

Consider an arbitrary object's motion in this graph. It has to accelerate each time its slope changes, but there is no curvature in the graph, which is a dead giveaway for acceleration in pos/time graphs. Its speed is constant here always, with the exception of at $t=0$, $t=2$, and $t=3.5$ (roughly). Its change in speed also seems nearly instant. I know this is an idealized object, and real motion would never be able to do this (no object can go from $5 \ m/s$ to $-15/3.5 \ m/s$ (roughly) instantaneously), but this kind of motion graph always puzzled me. It's taught pretty early on in kinematics yet it is glaringly unrealistic and unintuitive, I find, despite its simplicity.

• What do you think actually happens at 2s? – tfb Aug 4 '17 at 7:07
• @tfb I think its velocity changes from that of a positive one to that of a negative one instantly, in, for lack of a better word, time $dt$. It had a large change in velocity in a hugely tiny amount of time. – sangstar Aug 4 '17 at 7:11
• then I think you need to work on your physical intuition. This is not meant to be as rude as it sounds: intuition is important, and since you've asked this question you obviously have it. What it should be telling you is that this can't happen: this picture describes something not physically possible. In fact, around 2s, if you looked closely, there would be a finite but high acceleration -- perhaps the object has been hit with a hammer or something. So around 2s the graph is smooth: the 1st derivative of position is always continuous: it must be a $C^1$ function of time. – tfb Aug 4 '17 at 8:04
• @tfb No offensive taken, however my gut is telling me this system is impossible and cannot happen, as I said the object cannot make jumps in its velocity instantaneously. I was just wondering if my thinking there was wrong. – sangstar Aug 4 '17 at 8:25
• No: you are completely correct. If I taught physics I would have strong objections to graphs like this, as I think they actively damage intuition: half or more of the problems people have with the twin paradox in relativity is down to exactly this sort of misleading picture. – tfb Aug 4 '17 at 8:28

## 1 Answer

You are quite correct that as the graph is drawn the velocity is discontinuous at $2$ and $3.5$ seconds so the acceleration $dv/dt$ is undefined there.

But as you say this sort of graph tends to be used in elementary courses where we don't usually go into that level of detail. If challenged I would simply say the acceleration is so large that you can't see the curved region of the graph at this scale. That is the velocity change is effectively instantaneous.

• What do you mean by "instantaneous acceleration"? – ccorbella Aug 4 '17 at 7:55
• @ccorbella: sorry my wording was a bit unclear. Does it make more sense now? – John Rennie Aug 4 '17 at 8:08
• Yes, thanks. In fact you are saying the same, but now I get it. Sorry, it must be just me. xD – ccorbella Aug 4 '17 at 8:11