Q.Galileo's law of odd numbers: "The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, $1 : 3 : 5 : 7 \ldots$]." Prove it.

The key idea is Uniformly Accelerated Motion (UAM), specifically free fall where the initial velocity $u=0$ and acceleration $a=g$ is constant. The distance $s$ covered in time $t$ is given by the kinematic equation $s = ut + \frac{1}{2}at^2$, which simplifies to $s = \frac{1}{2}gt^2$ for a body falling from rest.

Let $T$ be the duration of one equal time interval.

1. The distance covered in the first interval ($t=T$) is $s_1 = \frac{1}{2}gT^2$.
2. The total distance covered from rest in the first two intervals ($t=2T$) is $S_2 = \frac{1}{2}g(2T)^2 = 4 \left(\frac{1}{2}gT^2\right) = 4s_1$. The distance covered only in the second interval is $d_2 = S_2 - s_1 = 4s_1 - s_1 = 3s_1$.
3. The total distance covered from rest in the first three intervals ($t=3T$) is $S_3 = \frac{1}{2}g(3T)^2 = 9 \left(\frac{1}{2}gT^2\right) = 9s_1$. The distance covered only in the third interval is $d_3 = S_3 - S_2 = 9s_1 - 4s_1 = 5s_1$. …

Galileo's law of odd numbers is a beautiful illustration of uniformly accelerated motion. It describes a specific pattern in how an object accelerates when falling freely, assuming air resistance is negligible. The core idea is to track the distance covered not just over a total time, but specifically within each successive, equal segment of time.

When a body falls "from rest," it means its initial velocity is zero. The acceleration acting on it is due to gravity, which we denote as $g$. For practical purposes near the Earth's surface, $g$ is considered constant. This scenario perfectly fits the definition of uniformly accelerated motion.

We will use this equation to calculate the distances covered. Since the body falls from rest, the initial velocity $u = 0$. The acceleration $a = g$. So, the equation simplifies to $s = \frac{1}{2}gt^2$.

Let's break down the proof step-by-step:

1. Define the equal time interval:

   Let $\Delta t$ be the duration of each equal time interval. So, the first interval is from $t=0$ to $t=\Delta t$, the second from $t=\Delta t$ to $t=2\Delta t$, and so on.

2. Calculate the distance covered in the first interval ($d_1$):

   The body starts from rest ($u=0$) at $t=0$. The time elapsed for the first interval is $t_1 = \Delta t$.

   The total distance covered from $t=0$ to $t=\Delta t$ is:

Let's call this fundamental distance $K = \frac{1}{2}g(\Delta t)^2$. So, $d_1 = K$.

3. Calculate the distance covered in the second interval ($d_2$):

To find the distance covered in the second interval (from $t=\Delta t$ to $t=2\Delta t$), we first find the total distance covered from $t=0$ to $t=2\Delta t$.

Total time elapsed is $t_2 = 2\Delta t$.

Total distance from $t=0$ to $t=2\Delta t$ is $S_2$:

The distance covered *in* the second interval, $d_2$, is the total distance up to $2\Delta t$ minus the total distance up to $\Delta t$:

4. Calculate the distance covered in the third interval ($d_3$): Similarly, for the third interval (from $t=2\Delta t$ to $t=3\Delta t$), we find the total distance covered from $t=0$ to $t=3\Delta t$. Total time elapsed is $t_3 = 3\Delta t$. Total distance from $t=0$ to $t=3\Delta t$ is $S_3$:

The distance covered *in* the third interval, $d_3$, is the total distance up to $3\Delta t$ minus the total distance up to $2\Delta t$:

> [!WARNING]
> A common mistake is to confuse the *total distance* covered from the start with the *distance covered in a specific interval*. Galileo's law refers to the latter. Always subtract the distance covered in previous intervals to find the distance in the current interval.

5. Generalize for the $n$-th interval ($d_n$): …

Concept: Average Velocity = Midpoint Velocity, for Linear $v(t)$

Method: Midpoint-Velocity Shortcut (no expand-and-subtract algebra)

For a body falling from rest, $v(t) = gt$ increases linearly with time. A general property of any linear function is that its average value over an interval equals its value at the interval's midpoint. Applying that single fact directly to each equal time slot proves the odd-number law in one line — no need to compute cumulative distances $S_n$ and subtract $S_{n-1}$ from $S_n$.

Steps

1. Set up equal intervals. Let $\Delta t$ be the duration of one interval; the $n$-th interval runs from $t=(n-1)\Delta t$ to $t=n\Delta t$.

2. Invoke the midpoint property. Since $v(t)=gt$ is linear in $t$, the average velocity during the $n$-th interval equals $v$ evaluated at the interval's midpoint, $t_{\text{mid}} = \left(n-\tfrac{1}{2}\right)\Delta t$:

3. Distance in the $n$-th interval = average velocity $\times$ duration:

4. Read off the ratio directly — the bracket $(2n-1)$ is the only part that depends on $n$; everything else, $\dfrac{g(\Delta t)^2}{2}$, is the same constant $K$ for every interval: $d_n = K(2n-1), \qquad K = \frac{g(\Delta t)^2}{2}$ …