If a particle travels 30 meters in any 3 seconds, does it necessarily travel 20 meters in any 2 seconds? Is it possible if its motion could've been discontinuous?
I'm trying to understand if there exists a function that has this property, but I chose to say it in terms of motion because it's easy to explain.
 A: This (30m in 3sec and 20m in 2sec) will only be applicable for motion with constant speed (ie. Linear function)
But if the motion is accelerated (ie. Degree of function is more than 1)
Or there are jerks in motion (ie. Function is discontinuous at some point(s))
then covering the given distance is possible in other ways.
For example when a person is traveling by car and reaches the destination in some time t then it is possible that he may have stopped the car somewhere or slowed due to traffic and may have had more speed at some other points. While another person travelled the same distance at constant speed and also reached at the same time t.
A: It isn't true, not even for continuous motion.
For example, let $V(t)=10+\cos\left(\dfrac{2\pi}{3}t\right)$ in meters/sec.
Then
\begin{eqnarray}
D&=&\int_0^3V(t)\,dt\\
&=&\left[10t+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}t\right)\right]_0^3\\
&=& 30
\end{eqnarray}
but
\begin{eqnarray}
D&=&\int_0^2V(t)\,dt\\
&=&\left[10t+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}t\right)\right]_0^2\\
&=& 20+\frac{3}{2\pi}\sin\left(\dfrac{4\pi}{3}\right)\\
&=&20-\frac{3\sqrt{3}}{4\pi}
\end{eqnarray}
For any three second time interval the distance traveled will be 30 meters because of the period of the sinusoidal portion of the velocity.
ADDENDUM: In response to a comment from @TheRubberDuck
\begin{eqnarray}
D&=&\int_x^{x+3}V(t)\,dt\\
&=&\left[10t+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}t\right)\right]_x^{x+3}\\
&=& 10(x+3)+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}(x+3)\right)-
\left(10x+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}x\right)\right)\\
&=&30+\frac{3}{2\pi}\left[ \sin\left(\frac{2\pi}{3}x+2\pi \right)-\sin\left(\dfrac{2\pi}{3}x\right)  \right]\\
&=&30
\end{eqnarray}
However, integrating between $x$ and $x+2$ gives a factor
\begin{eqnarray}
 \sin\left(\frac{2\pi}{3}x+\frac{4\pi}{3} \right)-\sin\left(\dfrac{2\pi}{3}x\right) 
\end{eqnarray}
which is not identically $0$.
A: No. Imagine a body that is stationary for two seconds, then moves $30$ meters in the next second, and then repeats this indefinitely. Then in any three-second interval, it moves $30$ meters, but there are clearly two-second intervals in which it doesn't move $20$ meters (in which, in fact, it doesn't move at all).
$$
x(t) = \begin{cases}
    30k & 3k \leq t \leq 3k+2 \\
    30k+30[t-(3k+2)] & 3k+2 \leq t \leq 3k+3
\end{cases} \qquad k \in \mathbb{Z}
$$
I've given a function that isn't differentiable, but it could be made differentiable quite easily.
A: What does it mean for the motion to be discontinuous?
If you are talking about the distance-time graph, it cannot be discontinuous.
${\text{The main thing}}$:
If a body travels $30$ meters in any $3$, it is not necessary that it travels $20$ meters in any $2$ seconds. For example, a body travels $5$ meters in the first second, $10$ in the second and $15$ in the third. Then $5$ again, then $10$ again, then $15$ again and so on.
See this:
$$5,10,15,5,10,15,\cdots$$
A: Suppose you have a vehicle traveling 10 m/s, and suppose the wheels have a diameter such that they make a complete revolution every three seconds. Then after three seconds, every point on the wheels will be at the same position relative to the vehicle, but will have moved 30m. On the other hand, after two seconds, points on the wheels will be in a different position, so they will have moved 20m plus or minus whatever distance they traveled relative to the vehicle.
