Problem 8 - Master Resource Book in JEE Main Physics

Question

A body starts from rest and moves with a constant acceleration. The ratio of distance covered in the \(n\)th second to the distance covered in \(n\) second is (a) \(\frac{2}{n}-\frac{1}{n^{2}}\) (b) \(\frac{1}{n^{2}}-\frac{1}{n}\) (c) \(\frac{2}{n^{2}}-\frac{1}{n}\) (d) \(\frac{2}{n}+\frac{1}{n^{2}}\)

Step-by-Step Solution

Answer

The correct answer is option (a) \( \frac{2}{n} - \frac{1}{n^2} \).

Step 1: Understanding the Problem

We are given a body that starts from rest and moves with constant acceleration. We need to find the ratio of distance covered in the \(n\)th second to the distance covered in \(n\) seconds. The options provided represent different mathematical expressions.

Step 2: Formula for Distance in nth Second

The distance covered in the \(n\)th second is given by the formula \( S_n = u + \frac{a}{2}(2n-1) \). Since the body starts from rest, \( u = 0 \), so \( S_n = \frac{a}{2}(2n-1) \).

Step 3: Total Distance in n Seconds

The total distance covered in \(n\) seconds is \( S = \frac{1}{2} a n^2 \) because the initial velocity \( u = 0 \) and the formula for total distance is \( S = ut + \frac{1}{2}at^2 \).

Step 4: Ratio of Distances

The problem asks for the ratio of distance covered in the \(n\)th second to the total distance covered in \(n\) seconds. We set up the ratio as: \[ \text{Ratio} = \frac{S_n}{S} = \frac{\frac{a}{2}(2n-1)}{\frac{1}{2} a n^2} \].

Step 5: Simplifying the Ratio Expression

By simplifying, we have: \( \text{Ratio} = \frac{2n-1}{n^2} \). We can split this expression into two fractions: \( \frac{2}{n} - \frac{1}{n^2} \), which matches option (a).

Step 6: Conclusion and Verification

The expression we derived for the ratio matches option (a) \( \frac{2}{n} - \frac{1}{n^2} \). Thus, option (a) is the correct answer.

Key Concepts

Distance-Time Relationship

In physics, understanding how distance and time relate to each other is crucial, especially when an object moves with constant acceleration. The distance a body covers over time is not always straightforward. When an object starts moving from rest, its initial velocity is zero. In this case, the relationship between distance and time can be described using the formula: \[ S = ut + \frac{1}{2}at^2 \]where:

• \(S\) is the total distance covered,
• \(u\) is the initial velocity,
• \(a\) is the constant acceleration,
• \(t\) is the time elapsed.

If an object begins from a standstill, the formula simplifies to \( S = \frac{1}{2}at^2 \).
This equation tells us that the distance covered by the object is proportional to the square of the time elapsed, which is a hallmark of motion under constant acceleration. This quadratic relationship implies that as time increases, the distance the object covers grows faster, making it an essential concept for solving various motion problems.

Nth Second Formula

The concept of the 'nth second' in the context of motion refers to the specific distance covered by an object in just the nth second of its motion. This is slightly different from considering the total distance covered over n seconds. The formula for finding the distance covered in the nth second is:\[ S_n = u + \frac{a}{2}(2n - 1) \]Since the object in our example starts from rest, \(u = 0\), simplifying the formula to:\[ S_n = \frac{a}{2}(2n - 1) \]This formula computes the distance the object travels during a specific second, accounting for constant acceleration.
For instance, if you want to know how much ground the object covers specifically in the 5th second, this formula helps us find that specific value, distinguishing it from the total ground up to that point. It's an essential tool for deeper insights into kinematics, clarifying how acceleration affects motion over discrete time intervals.

Kinematic Equations

Kinematic equations are the cornerstone of classical mechanics for motion analysis under constant acceleration. They enable us to predict and describe the motion of objects without having to understand the underlying forces at a given time. There are a set of four key kinematic equations:

• \(v = u + at\)
• \(S = ut + \frac{1}{2}at^2\)
• \(v^2 = u^2 + 2aS\)
• \(S = \frac{(u + v)}{2}t\)

These equations correspond to different aspects of motion:

• Velocity changes due to constant acceleration,
• Distance covered over time with constant acceleration,
• Relating initial velocity, final velocity, and distance covered,
• The average velocity if acceleration is constant.

These tools equip us to solve a variety of problems involved in constant acceleration, such as the one outlined in the exercise. They reveal how initial conditions like initial speed and time relate to final outcomes such as total distance and end speed. By mastering these, one learns to navigate the dynamic relationships that govern real-world movements.