What is the rate of change of time wrt velocity of an object? disclaimer, I'm just an average highschooler so please be a little friendly with the mathematics of your answers but I wondered what would be $dt/dv$?
 A: In straight line motion, if the velocity is increasing we can not only find the velocity for a given time, but we can also find the time for a given velocity. In this case we can say that time is a function of velocity, and we can work out what $\frac{dt}{dv}$ is.
It can be shown mathematically that $\frac{dt}{dv}$ is $1/\frac{dv}{dt}$, in other words $\frac{1}{\text{acceleration}}$.
We can do the same thing if the velocity is decreasing.
If the velocity increases and decreases there will be two or more time ranges for some velocity ranges. You would need to restrict yourself to discussing only one of the ranges.
A: $\frac{dt}{dv}$ tells you how long it takes to change your velocity. For example suppose I'm driving my car at $60$ mph and I want to know how long it will take me to accelerate to $65$ mph then this time is given by (assuming $\frac{dt}{dv}$ is constant):
$$ \Delta t_{60\to65} = \frac{dt}{dv} \times 5 $$
If $\frac{dt}{dv}$ is not constant then this becomes an integral:
$$ \Delta t_{v_1\to v_2} = \int_{v_1}^{v_2} \frac{dt}{dv} dv $$
Although derivatives are not fractions we physicists often treat them as if they are and get away with it. If you look at it this way you'll see that $\frac{dt}{dv}$ multiplied by a change in velocity $dv$ gives you a change in time $dt$. More formally for a small change in velocity $\delta v$:
$$ \delta t \approx \frac{dt}{dv} \delta v $$
A: To answer this question, one needs to briefly look at what dependent and independent variables are. Mathematically, a dependent variable $y$ is defined as a function of a (say, finite, for simplicity) set of independent variables $\{x_i\}$. In other words,
$$
\begin{equation}
 y = f(x_1, x_2, \dots x_n)
\end{equation}
$$
Let us now drop down to functions of a single variable. For given a set of times $t$ and velocities $v$, it should be obvious which one is dependent and which is independent $(v \equiv v(t))$. In calculus, we typically take derivatives of these functions with respect to the independent variables and hence things like $a = \frac{d}{dt}(v(t))$ aka acceleration make sense.
Of course, since this is mathematics in physics after all, my functions are usually very nice so I would probably be able to define a $t \equiv t(v)$ by inverting the velocity function and then we could have a rate of change of time wrt velocity as well, which is very hard to wrap one's head around for good reasons. In other news, you might also see the derivative being treated as a fraction sometimes in physics and that might suggest $\frac{dt}{dv} = a^{-1}$ (and that somehow works out fine dimensionally).
