So, again my over-curiosity arose a problem for me, this time in Physics class. Being able to solve the questions sir gave me before the time limit, I was told to solve another, a little tougher.

This is the exact words of my Physics Teacher : "A particle of unit mass undergoes $1$-dimensional motion such that its velocity expressed as a function of its position is: $$V(x) = bx^{-2n}$$

where, $b$ and $n$ are constants, and $x$ denotes the position of the particle. Prove that, Acceleration $a$ varies with $x$."

Now, I spent a lot of time and came up with this :

$$a = \frac{dv}{dt} = \left(\frac{dv}{dx}\right)\left(\frac{dx}{dt}\right)$$

By definition, $\frac{dx}{dt} = V(t)$. But Velocity is expressed as a function of position, not time.

Despite that, I went ahead and put $V(x)$. That gave me $$a = \frac{dv}{dt} = \left(\frac{dv}{dx}\right)\left(\frac{dx}{dt}\right)$$

$$= \frac{d}{dx}(bx^{-2n}) * V(x)$$

$$= -2nbx^{-2n - 1} * bx^{-2n}$$

$$= -2nb^2x^{-4n - 1}$$

My teacher said I was correct. But what about $V(t)$? The function I put into calculation is $V(x)$, not the $V(t)$ we're supposed to put normally.

Is $V(x) = V(t)$? If it is, How can we prove it?

How can we proceed with this?

  • 1
    $\begingroup$ Great question. It's likely that the position depends on time. So really what you have is the velocity as a function of position, which is a function of time: $V(x(t))$. If you knew $x(t)$ explicitly, you could just plug it in to get $V(t)$. $\endgroup$ – zbrads2 May 19 at 3:03
  • $\begingroup$ Exactly. I wasn't informed anything regarding $V(x(t))$ here. $\endgroup$ – Soumalya Pramanik May 19 at 3:14
  • 1
    $\begingroup$ Sure, but the point is that $V(x)$ is the same thing as $V(t)$ up to a change of variables. So it is still true that the velocity is $\frac{dx}{dt}$ even though it's expressed as a function of position. $\endgroup$ – zbrads2 May 19 at 3:45
  • $\begingroup$ I thought so! Both of them vary with time, and $x(t)$ represents position at that instant, and $V(t)$ represent velocity at the same instant. So, We can still write $V(x) = V(t)$! YES! Thanks for the help, Now I understand why the answer was correct, despite my method being mathematically wrong $\endgroup$ – Soumalya Pramanik May 19 at 3:48

By definition, velocity is $\frac{dx}{dt},$ the rate of change of position relative to the change in time, also called the derivative of position with respect to time. There is nothing in that definition that says velocity is a function of anything.

We determine what velocity is a function of by seeing what properties we can track for which there is a uniquely identified velocity at each value of the property. Since we assume position is a function of time (the same object cannot be in two places simultaneously), and velocity is the derivative of position with respect to time, we can also work out velocity as a function of time. But if the object travels through each position on its path just once and never returns, it has only one velocity at each position, and therefore there is a perfectly well-defined function from positions on the path to velocities of the object.

There are many cases in which velocity is not a function of position, for example the swinging of a pendulum, which can have two exactly opposite velocities at a single point. But in your question you have a velocity that is always positive, so the particle just keeps traveling in one direction and never stays at a point or comes back to visit the same point again. So velocity as a function of position makes perfect sense.

You can call the function from position to velocity $V(x)$. Mathematically, however, that is a different function than the function from time to velocity, and it is an abuse of notation (and potentially confusing, as you found out) to use the same name for two different functions. So if you use $V(x)$ for the function from position to velocity, it is better not to say that $V(t)$ is velocity as a function of time. Better come up with some other name.

We could say that $V_t$ is the name of the function from time to velocity, that is, the velocity at time $t$ is $V_t(t).$ Then with $V(x)$ defined as in your question, if $x(t)$ is the position at time $t,$ then $V(x(t)) = V_t(t),$ that is, both sides of the equation tell the velocity the particle had as it passed through position $x(t)$ at time $t$.

Again, in the fact that by definition $a = \frac{dv}{dt}$ and the fact that according to the chain rule, $\frac{dv}{dt} = \left(\frac{dv}{dx}\right)\left(\frac{dx}{dt}\right),$ it is not necessary to say that any of the quantities $a,$ $\frac{dv}{dt},$ $\frac{dv}{dx},$ or $\frac{dx}{dt}$ is a function of anything in particular. It is just enough to know that under some particular circumstances these quantities all have unique values, and the values then must satisfy these equations. Very often, if we know (or have deduced) the position $x$ as a function of time, it is convenient to treat all these things as functions of time, but it is not always necessary. Since in your problem all these things are just as much functions of position as functions of time, your method is fine.

  • $\begingroup$ Exact same way I interpreted the problem. Thanks for clarifying it even more! $\endgroup$ – Soumalya Pramanik May 19 at 4:03

The notation is a bit confusing.

$V(t)$ represents the velocity of the particle at a particular time $t$. In some cases the position of the particle has a one-one relationship with the time, in these circumstances we can describe different moments of the motion in terms of the position rather than the time. This means that there is a different function $V(x)$ which descibes the velocity of the particle when it is at the position $x$. These two functions are related by the fact that when $x(t)$ is substituted into $V(x)$ we get $V(t)$.

To see this lets consider a few specific examples.


$$V(x) = Cx,$$

where $C$ is some constant. This implies the differential equation,

$$ \frac{dx}{dt} = C x,$$

which is solved by,

$$ x(t) = Ae^{Ct},$$

where $A$ is another constant. Substituting this into our expression for $V(x)$ we get,

$$ V(t) = CA e^{Ct},$$

which is the same result we would have gotten by computing $x'(t)$.

Consider a particle moving to the right with a constant acceleration $a$,

$$x(t)= \frac12 at^2,$$

then we can compute by differentiation that,

$$V(t) = at,$$

but its also true that $$ t = \sqrt{2\frac{x}{a}}$$,

and so we have,

$$ V(x) = a \sqrt{2\frac{x}{a}}$$

  • $\begingroup$ Exactly. The functions are different - But, They give the same values for $V(x)$ and $V(t)$ as $x$ is the position at time $t$ and $V(t)$ is the velocity at time $t$. Which allows us to write $V(x) = V(t)$. $\endgroup$ – Soumalya Pramanik May 19 at 4:01

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