A small block slides without friction down an inclined plane starting from rest. Let s_n be the dist

Q: Wait, why can't we just use $s = ut + \frac{1}{2}at^2$?

You absolutely can! The distance in the $n^{\text{th}}$ second is simply the distance covered in $n$ seconds minus the distance covered in $(n-1)$ seconds. If you calculate $\frac{1}{2}an^2 - \frac{1}{2}a(n-1)^2$, you will get exactly $\frac{a}{2}(2n-1)$. The formula is just a time-saving shortcut!

Q: Why is the initial velocity $u$ zero for both $s_n$ and $s_{n+1}$?

The formula $S_n = u + \frac{a}{2}(2n - 1)$ always uses the *absolute* initial velocity at $t=0$. Even when calculating the distance for the $(n+1)^{\text{th}}$ second, $u$ refers to the velocity at the very beginning of the motion, not the velocity at $t=n$.

Q: Does the angle of the inclined plane matter?

The angle determines the value of the acceleration ($a = g \sin\theta$). However, since we are taking a ratio of distances, the acceleration $a$ cancels out completely. So, the ratio is independent of the incline's angle!

Q: What if the block had an initial velocity?

If $u \neq 0$, the $u$ term would not vanish, and the ratio would depend on both $u$ and $a$. The beautiful cancellation of $a/2$ only happens because the block starts from rest.

Q: Is this related to Galileo's law of odd numbers?

Yes! Galileo discovered that for an object starting from rest under constant acceleration, the distances covered in equal successive time intervals are in the ratio of odd numbers: 1 : 3 : 5 : 7... If you plug in $n=1, 2, 3$ into our formula, you'll see exactly this pattern!

Question

A small block slides without friction down an inclined plane starting from rest. Let $s_n$ be the distance travelled from $t = n - 1$ to $t = n$. Then, $\frac{s_n}{s_{n + 1}}$ is

Options:

(A) $\frac{2n - 1}{2n}$

(B) $\frac{2n + 1}{2n - 1}$

(C) $\frac{2n - 1}{2n + 1}$

(D) $\frac{2n}{2n + 1}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a small block placed at the top of a frictionless inclined plane. The moment you let it go, gravity pulls it down. Because there is absolutely no friction to oppose its motion, the block accelerates downwards at a constant rate.

Let's call this constant acceleration $a$. The block starts from rest, which is a crucial piece of information. It means that at time $t = 0$, the initial velocity $u$ is exactly $0$.

### The Master Equation

We are asked to find the ratio of distances traveled in two specific, consecutive one-second intervals. To do this efficiently, we need to pull a very specific tool from our kinematics arsenal: the formula for the distance traveled in the $n^{	ext{th}}$ second.

The distance covered by an object in the $n^{	ext{th}}$ second of its motion is given by:
$$S_n = u + \frac{a}{2}(2n - 1)$$

This formula is incredibly powerful because it saves us from having to calculate the total distance at $t=n$ and subtract the total distance at $t=n-1$. It gives us the interval distance directly!

### Calculating the First Interval

The problem defines $s_n$ as the distance traveled from $t = n - 1$ to $t = n$. This is exactly the definition of the distance traveled in the $n^{	ext{th}}$ second.

Since our block started from rest, we can substitute $u = 0$ into our master equation.

$$s_n = 0 + \frac{a}{2}(2n - 1)$$

This simplifies beautifully to:
$$s_n = \frac{a}{2}(2n - 1)$$

### Calculating the Second Interval

Next, we need to find $s_{n+1}$, which is the distance traveled in the very next second, from $t = n$ to $t = n + 1$.

To find this, we simply take our expression for $s_n$ and replace every instance of $n$ with $(n+1)$.

$$s_{n+1} = \frac{a}{2}(2(n+1) - 1)$$

Let's expand the terms inside the parenthesis. Distributing the $2$ gives us $2n + 2$. Subtracting $1$ leaves us with $2n + 1$.

$$s_{n+1} = \frac{a}{2}(2n + 1)$$

### The Final Calculation

We now have clean expressions for both $s_n$ and $s_{n+1}$. The final step is to find their ratio.

$$\frac{s_n}{s_{n+1}} = \frac{\frac{a}{2}(2n - 1)}{\frac{a}{2}(2n + 1)}$$

Notice how the constant factor $\frac{a}{2}$ appears in both the numerator and the denominator. Because the acceleration is constant and non-zero, we can completely cancel this term out!

**$$\frac{s_n}{s_{n+1}} = \frac{2n - 1}{2n + 1}$$**

This elegant result shows that the ratio of distances in consecutive seconds for an object starting from rest is purely a function of time, completely independent of the actual acceleration!

Animated Solution for Physics - Kinematics: A small block slides without friction down an inclined plane starting from rest. Let $s_{n}$ be the distance travelled from $t = n - 1$ to $t = n$ . Then, $\frac{s _{n}}{s _{n + 1}}$ is

Select Answer:

Visualized Solution

Visualizing the Motion

The $n^{th}$ Second Formula

Setting up $s_{n}$

Computing $s_{n}$

Setting up $s_{n + 1}$

Computing $s_{n + 1}$

The Final Ratio

The Sigma Insight: Motion in a Straight Line

Analyzing the Setup

The Master Equation

Calculating the First Interval

Calculating the Second Interval

The Final Calculation

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