Animated Solution for Physics - Kinematics: A particle covers a distance unidirectionally with uniform acceleration. If its average velocity is $v_{\text{av}}$, what could be range of modulus of its instantaneous velocity at the midpoint of the path?

It covers a total distance $d$. Let's say it starts with an initial velocity $u$ and finishes the journey with a final velocity $v_f$. We are interested in finding the instantaneous velocity $v$ exactly at the midpoint of this path, which is at a distance of $d/2$ from the start.

Because the acceleration is uniform, we have a very neat property for the average velocity $v_{\text{av}}$ over the entire trip. It is simply the arithmetic mean of the initial and final velocities:

How fast is the particle going when it's exactly halfway there? We can use the third equation of motion for the first half of the journey:

We don't know the acceleration $a$ or the distance $d$, so let's eliminate them. If we apply the same third equation to the entire journey, we get:

Let's substitute this value of $ad$ back into our midpoint velocity equation:

Taking the square root, we find that the midpoint velocity is the root mean square (RMS) of $u$ and $v_f$:

For any two numbers, their Root Mean Square is always greater than or equal to their Arithmetic Mean. Since our midpoint velocity is the RMS, and our average velocity is the AM, it directly implies that the midpoint velocity $v$ must be greater than or equal to the average velocity. That's our lower bound!

Now, what about the upper limit? Let's rewrite the numerator $u^2+v_f^2$ as $(u+v_f{)}^2-2u{v}_f$. We know $u+v_f$ is twice the average velocity ($2v_{\text{av}}$). So, $v^2$ becomes:

To make $v$ as large as possible, we need to subtract the smallest possible value. The problem states the motion is unidirectional. This means the particle never turns back, so $u$ and $v_f$ must have the same sign (let's say they are both positive). The smallest possible value for their product, $u{v}_f$, is zero. This happens if the particle starts from rest ($u=0$) or comes to a stop ($v_f=0$).

Plugging in zero, the maximum value for $v^2$ is $2v_{\text{av}}^2$. Taking the root gives us our upper bound:

By combining our lower bound from the RMS-AM inequality and our upper bound from the unidirectional constraint, we find the exact range for the instantaneous velocity at the midpoint:

$$v_{\text{av}}\le v\le \sqrt{2}{v}_{\text{av}}$$

A brilliant mix of kinematics and pure algebra!