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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?

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    $\begingroup$ 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 at 10:03
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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 $y'$.

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