# Velocity-Time graph confusion

I'm not 100% sure whether this is best answered with regards to Maths or Physics, but I feel this is more mathematical.

If you take the equation $y = 40-10x$. If you change $y$ to $v$ (velocity in $ms^{-1}$), and $x$ to $t$ (time in $s$), you can use this to model the velocity of a ball, thrown upwards at $40ms^{-1}$ (to keep things simple, I am assuming the acceleration due to gravity is $10ms^{-2}$ and ignoring air resistance).

Now, if I make $v=0$, then $0=40-10t \rightarrow 10t=40 \rightarrow t=4$. So, at exactly 4 seconds, the velocity is $0$. This makes total sense to me.

However, if I rearrange the equation to be in terms of $t$, I get $10t=40-v \rightarrow t=4-\frac{v}{10}$. Now, if I make $t$ equal to $4+h$, I get $4+h=4-\frac{v}{10} \rightarrow h=\frac{-v}{10} \rightarrow v=-10h$. As you can see, as long as h isn't $0$, regardless of how small $h$ is, if $t=4+h$, $v$ will not be $0$.

If we go back and remember that this is a ball that has been thrown upwards, then obviously it must "stop" shortly before falling back down. However, as I have shown, it stops for absolutely no time at all.

What is confusing me, is that both of the below statements are true;

1. The ball has a velocity of $0$ for no time at all, and if an action takes up $0$ time, then surely this action never happened. However;
2. As I say above, we can clearly prove that the velocity must have been $0$ at one point given it changes direction.

Surely these 2 statements contradict each other. Can anyone explain why it is possible for both to be true?

• When $h=0$, the ball is at maximum height. – Karn Watcharasupat Apr 12 '18 at 10:09
• So, essentially your question is about the duration of a speed. But the speed, defined by $v = \frac{dx}{dt}$, is an instantaneous quantity itself. If you threw an actual ball in real life and measured its velocity with infinite accuracy, you would get something that is not exactly a straight line. If we set the velocity of an object to be a function of time, it's only a model of real life and not 100% accurate. – Matti P. Apr 12 '18 at 10:13
• Statement $1$ is not well posed, the ball has velocity $v=0$ exactly at $t=4$ as it has $v=-10$ at $t=5$ and so on. – gimusi Apr 12 '18 at 10:15
• @gimusi - The point I was trying to make in statement 1 is that $v=0$ only at one point, and the time for which it equals $0$ is $0$ seconds. One could argue that $v=0$ between the times of $4$ seconds and $4+h$ seconds (or in other words it is $0$ for $h$ seconds). However, unless $h=0$, you could simply make $h$ smaller and smaller (eventually reaching $0$). However, if $h=0$, then as $v$ is $0$ for $h$ seconds, then surely $v$ was never equal to $0$ if it was only equal to it for $0$ seconds. This is what is confusing me with this situation. – PhysicsGuy123 Apr 12 '18 at 11:34

The result is consistent indeed

• when $h=0 \implies v=0$

• when $h>0 \implies v<0$ since the ball is falling down

• I was not questioning the consistency of the maths. My point was that you could make h infinitesimally small, but $v$ still wouldn't equal $0$. Therefore, if $v$ can only be equal to $0$ at one instant (when $h=0$), the time for which it is $0$ is no time at all, so my question is how can something happen if the action took no time. – PhysicsGuy123 Apr 12 '18 at 11:24
• If you use a mathematical continuous model to describe the phenomenon there is not inconstency in it. In this model v=0 exactly for the instant t=4. – gimusi Apr 14 '18 at 14:12

"... if an action takes up $0$ time, then surely this action never happened ..."

This is where your error lies. In classical physics we assume that space and time are continuous and can be divided into parts that are as small as we like. Under these assumptions it is possible for an event to have zero duration i.e. to occur at a single instant in time. It is also possible for an event to have zero spatial extent i.e. to occur at a single point in space.

If you have philosophical doubts about these assumptions then you may be reassured by the fact that we know that classical physics is only an approximation to reality. In quantum physics, Heisenberg's uncertainty principle prevents an event having zero duration or zero extension - there is always some unavoidable uncertainty over the time or location at which an event occurs.

But at the scale of throwing a ball, quantum physics has a negligible effect and so we use the methods of classical physics.

• Thank you for this explanation. One thing that confuses me is you say, "space and time are continuous and can be divided into parts that are as small as we like". This means you can divide time, $t$, into $x$ parts of size $p$, then you would have $\frac{t}{x}=p$, or $t=p \times x$. Given the size of the parts in my question are of size $0$, we would have $t=0 \times x \rightarrow t = 0$, so the pieces can only be equal to $0$ if $t$ also equals $0$, which it clearly doesn't. This therefore leads to another, similarly interesting, puzzle. – PhysicsGuy123 Apr 12 '18 at 11:58