Animated Solution for Physics - Kinematics: An engine of a train moving with uniform acceleration, passes the signal-post with velocity $u$ and the last compartment with velocity $v$. The velocity with which middle point of the train passes the signal post is

A Train Passing a Post

Imagine you are standing next to a railway track, watching a train pass by a signal post. The train is accelerating uniformly. When the very front of the engine passes you, it has a velocity of $u$. By the time the very last compartment passes you, the train has sped up to a velocity of $v$.

The question asks for the velocity of the exact midpoint of the train as it passes that same signal post. Let's call this unknown velocity $v^{\prime }$.

Instead of visualizing a moving train and a stationary post, it is mathematically identical (and often easier) to imagine a single particle accelerating over a distance $l$ (the length of the train). The particle starts at $x=0$ with velocity $u$, reaches $x=l/2$ with velocity $v^{\prime }$, and finally reaches $x=l$ with velocity $v$.

Kinematics Without Time

Since the acceleration $a$ is uniform and we are dealing with velocities and distances (without any mention of time), the perfect tool for the job is the third equation of motion:

This equation elegantly links the initial state, the final state, the acceleration, and the spatial displacement.

First, let's apply this equation to the entire length of the train. The initial velocity is $u$, the final velocity is $v$, and the total displacement is $l$. Plugging these into our master equation gives:

From this, we can easily isolate the term $2al$ by moving $u^2$ to the other side:

This gives us a direct expression for the acceleration in terms of the known velocities and the length of the train: $a=\frac{v^2-u^2}{2l}$.

Now, let's shift our focus to the first half of the train. The displacement is now exactly half the length, $s=\frac{l}{2}$. The initial velocity is still $u$, but the final velocity at this midpoint is our target variable, $v^{\prime }$. Applying the third equation of motion again:

Notice how beautifully the $2$ in the numerator and the denominator cancel out. This simplifies our equation to:

We already know from our full-train analysis that $2al=v^2-u^2$. Therefore, the term $al$ is simply half of that difference:

Let's substitute this back into our midpoint equation:

To add these terms, we take a common denominator of $2$:

Finally, taking the square root of both sides yields the velocity of the midpoint:

This result is profound. It tells us that for any object undergoing uniform acceleration, its velocity at the exact spatial midpoint is the Root Mean Square (RMS) of its initial and final velocities. It is a classic, symmetric, and highly testable property of kinematics!