While revising the topic, "Motion in a plane" from "Concepts of Physics by HC. Verma", I had the following question in my mind.

Question:~ When x- component of velocity of a particle is extremum provided that it travels at uniform velocity along x-direction in X-Y plane or in other words $a_x$(x-component of acceleration is $0$), i.e $\overrightarrow a = a_y \hat j$?

In order to answer this question, I've attempted as follows,

My answer:~ $\overrightarrow v = v_x \hat i + v_y \hat j$ ...(Velocity vector) $$\overrightarrow v = \frac{dx}{dt} \hat i + \frac{dy}{dt} \hat j$$ If we represent $|\overrightarrow x|$ as magnitude of $\overrightarrow x$, then

$$|\overrightarrow v| = v = \sqrt[]{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}$$ $$vdt= \sqrt[]{dx^2 + dy^2}$$ $$vdt = dx\left(\sqrt[]{1+y'^{2}}\right)$$ $$v = v_x \left ( \sqrt[]{1+y'^{2}}\right)$$ Here, $v_x = \frac{v}{\sqrt[]{1+y'^{2}}}$ is $x$-component of velocity. Now we need to make $v_x$ extremum with respect to x coordinate since we have information that particle travels at uniform velocity along x-direction, i.e $\frac{\partial v}{\partial x} = 0$. We set, $\frac{d v_x}{d x} = 0$ to make $v_x$ extremum. We then have, $$\frac{y"y'}{1 + y'^2} = 0$$

From, here three possibilities arise.

  1. $y" = 0$
  2. $y' = 0$
  3. $y' = Undefined$

My main query is as follows.

If we consider a projectile motion, which is a nice example of motion in the plane with constant acceleration, we know that at maximum height, $x$- component of velocity is maximum. In that case $\overrightarrow a = -g \hat j$. Hence, in this case, $y' = 0$ is a valid case of $x$-component of velocity to be maximum, and $y" = 0$ represents the case for which $x$-component is minimum at boundaries. But, then what $y' = undefined$ represents. Or more specifically, what is an example that includes $y' = undefined$ be the condition for x-component of velocity to be minimum?

  • 1
    $\begingroup$ If the x component of velocity is constant its maximum and minimum are equal to that constant value. This is independent of any motion in the y direction. The values of y, y' and y'' are not defined by the value of x; they must be defined separately. $\endgroup$
    – Peter
    Jul 28 '21 at 10:03

Using the word extremum for the x-component of velocity is misleading in this context, since you have stated that the x-component of velocity is constant. The statement that $v_x$ is constant immediately implies that any derivative of $v_x$ is zero. It also implies that $v_y$ and $y'$ are proportional, as $v_y=y'v_x$.

When you differentiate both sides of $v = v_x \left ( \sqrt[]{1+y'^{2}}\right)$ with respect to $x$ you should get $$v'=v_x \frac{y'y''}{\sqrt{1+y'^2}}$$ While you did not state explicitly that you are considering the situation where the speed is extremal, this is what you have done by putting $v'=0$. The following are the conditions when the speed is extremal.

If $y'=0$ then $v_y=y'v_x=0$; any change in $v_y$ will increase the speed.

If $y''=0$ then $v_y$ is a maximum or a minimum; any change in $v_y$ will change the speed.

In the limiting case where $\frac{1}{\sqrt{1+y'^2}}\to0$, we would have $v_y\to\infty$, but in this case $\frac{y'}{\sqrt{1+y'^2}}\to1$ and so we would get $v'/y''\to 1$. This would be a situation where the speed increases to infinity, but at a rate (acceleration) that decreases to zero. Note that we do not need to worry about the possibility of an imaginary solution because $\sqrt{1+y'^2}\ge 1$ for all real $y'$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.