JEE PYQ: Motion in a Straight Line - Question ID f1c758243f41 (JEE Main 2024)

Core Formula & Concept:

The problem deals with a body moving on a frictionless plane starting from rest. This implies the body undergoes uniformly accelerated motion (constant acceleration) along a straight line. The key formulas for such motion are:

• Displacement in the $n$-th second ($S_n$): The distance covered between time $t = n-1$ and $t = n$ is given by: $S_n = u + \frac{a}{2} (2n - 1)$ where $u$ is the initial velocity (here $u = 0$ since the body starts from rest), and $a$ is the constant acceleration.
• For a body starting from rest ($u = 0$), the formula simplifies to: $S_n = \frac{a}{2} (2n - 1)$
• Similarly, the distance covered in the $(n-1)$-th second is: $S_{n-1} = \frac{a}{2} [2(n-1) - 1] = \frac{a}{2} (2n - 3)$

The problem asks for the ratio $\frac{S_{n-1}}{S_n}$ for $n = 10$, expressed in the form $\left(1 - \frac{2}{x}\right)$. Our goal is to find the value of $x$.

Step-by-Step Derivation:

Step 1: Write expressions for $S_n$ and $S_{n-1}$

For a body starting from rest under constant acceleration $a$, the distance covered in the $n$-th second is: $S_n = \frac{a}{2} (2n - 1)$ Similarly, the distance covered in the $(n-1)$-th second is: $S_{n-1} = \frac{a}{2} (2n - 3)$

Step 2: Compute the ratio $\frac{S_{n-1}}{S_n}$

Taking the ratio: $\frac{S_{n-1}}{S_n} = \frac{\frac{a}{2} (2n - 3)}{\frac{a}{2} (2n - 1)} = \frac{2n - 3}{2n - 1}$ The acceleration $a$ and the factor $\frac{a}{2}$ cancel out.

Step 3: Substitute $n = 10$

For $n = 10$: $\frac{S_{9}}{S_{10}} = \frac{2(10) - 3}{2(10) - 1} = \frac{20 - 3}{20 - 1} = \frac{17}{19}$

Step 4: Express the ratio in the given form

The problem states that: $\frac{S_{n-1}}{S_n} = 1 - \frac{2}{x}$ Substituting the computed ratio: $\frac{17}{19} = 1 - \frac{2}{x}$

Step 5: Solve for $x$

Rearrange the equation: $\frac{2}{x} = 1 - \frac{17}{19} = \frac{2}{19}$ Thus: $x = 19$

Common Traps & Exam Tip:

1. Misidentifying the time intervals: Students often confuse $S_n$ as the total distance covered up to the $n$-th second rather than the distance covered during the $n$-th second. Remember, $S_n$ is the distance covered between $t = n-1$ and $t = n$.
2. Incorrect formula for $S_n$: The formula $S_n = u + \frac{a}{2} (2n - 1)$ is derived from the difference of displacements at $t = n$ and $t = n-1$. Forgetting the $(2n - 1)$ term or misapplying it leads to errors.
3. Algebraic mistakes in solving for $x$: When solving $\frac{17}{19} = 1 - \frac{2}{x}$, students sometimes incorrectly cross-multiply or misplace terms. Always isolate $\frac{2}{x}$ first.
4. Assuming non-uniform acceleration: The problem specifies a frictionless plane and the body starts from rest, implying constant acceleration. Do not overcomplicate the problem by considering variable acceleration.

Final Answer: The value of $x$ is 19.