Problem 24 - Master Resource Book in JEE Main Physics

Question

A particle moving in a straight line with uniform acceleration is observed to be a distance \(a\) from a fixed point initially. It is at distances \(b, c, d\) from the same point after \(n, 2 n, 3 n\) second. The acceleration of the particle is (a) \(\frac{c-2 b+a}{n^{2}}\) (b) \(\frac{c+b+a}{9 n^{2}}\) (c) \(\frac{c+2 b+a}{4 n^{2}}\) (d) \(\frac{c-b+a}{n^{2}}\)

Step-by-Step Solution

Answer

The acceleration is \( \frac{c-2b+a}{n^2} \), which is option (a).

Step 1: Apply the equation of motion

Let the initial velocity of the particle be \( u \) and the uniform acceleration \( a \). The position of the particle as a function of time can be described by the equation: \[ s = ut + \frac{1}{2} a t^2 \]Initially, the distance from the fixed point is \( s_0 = a \).

Step 2: Write distance expressions at given times

After \( n \) seconds, the distance from the fixed point is given by:\[ b = a + un + \frac{1}{2} a n^2 \]After \( 2n \) seconds, the distance is:\[ c = a + u(2n) + \frac{1}{2} a (2n)^2 = a + 2un + 2a n^2 \]After \( 3n \) seconds, the distance is:\[ d = a + u(3n) + \frac{1}{2} a(3n)^2 = a + 3un + \frac{9}{2} a n^2 \]

Step 3: Derive equations for comparison

Subtract the time \( n \) equations from each other to eliminate \( u \):From \( c - b \),\[ c - b = (a + 2un + 2a n^2) - (a + un + \frac{1}{2} a n^2) = un + \frac{3}{2} a n^2 \]From \( d - c \),\[ d - c = (a + 3un + \frac{9}{2} a n^2) - (a + 2un + 2a n^2) = un + \frac{5}{2} a n^2 \]

Step 4: Eliminate variable for acceleration

Set the expressions for differences \( c - b \) and \( d - c \) to establish equality for consistent acceleration:\[ un + \frac{3}{2} a n^2 = un + \frac{5}{2} a n^2 \]This equates to:\[ \frac{3}{2} a n^2 = \frac{5}{2} a n^2 \]

Step 5: Solve for acceleration

Reform the equation:\[ b = a + un + \frac{1}{2} an^2 \]\[ c = a + 2un + 2a n^2 \]\[ \Rightarrow c - 2b + a = un + \frac{3}{2} a n^2 - 2un - a n^2 + a = c - 2b + a = \frac{1}{2} an^2 \]Solve for acceleration:\[ a = \frac{c - 2b + a}{n^2} \]

Step 6: Identify correct option

Using the formula derived for acceleration, compare it with the given options to recognize that option (a) \( \frac{c-2b+a}{n^2} \) correctly represents the acceleration.

Key Concepts

Equation of Motion

When studying the motion of particles, the equation of motion is fundamental. It's used to describe how an object moves under certain conditions. In this problem, we're dealing with a particle moving in a straight line with consistent acceleration. The equation of motion is given by:\[ s = ut + \frac{1}{2} a t^2 \]Here, \( s \) represents the distance traveled by the particle, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time elapsed.
This formula is crucial because it helps us determine the position of the particle at any given time using its initial velocity and acceleration. We can see how each variable affects the particle's motion over time, which will be important in solving the original problem.

Initial Velocity

Initial velocity, often denoted as \( u \), is the velocity of the particle at the starting point of observation. It's a key component in calculating the particle's position at different times. In the context of the problem, the initial velocity allows us to determine the distance traveled over specific time intervals.
If the particle starts from rest, \( u \) would be zero. But, if it begins with some velocity, \( u \) will have a positive value. Knowing \( u \) is essential in using the equation of motion and understanding how much distance is covered initially before the particle speeds up due to acceleration.

Distance-Time Relationship

The distance-time relationship describes how the distance of the particle from the fixed point changes over time. In uniform acceleration, this relationship is quadratic, depicted by the equation of motion we’ve seen:\[ s = ut + \frac{1}{2} a t^2 \]This quadratic nature means the distance doesn't change linearly but increases more rapidly as time goes on due to the effect of the squared time term.
In the original problem, these distances were observed at time intervals of \( n \), \( 2n \), and \( 3n \). Each change in distance gives us information that can be used to establish the relationship between the distances at these times.
Understanding this relationship allows us to solve for variables like acceleration or initial velocity, using differences between the distances calculated at these specific times.

Particle Motion

Particle motion under uniform acceleration is an important concept in physics. It helps us understand how objects move when subjected to constant forces. In our scenario, the particle starts at a known distance from a fixed point and covers different distances over time.
Analyzing the motion involves looking at how the initial velocity and acceleration affect the particle's trajectory. When looking at how far the particle travels, we consider both the initial speed it had and how acceleration increases that speed over time.

• Initial Position: The known starting point or distance from the fixed point.
• Velocity Changes: Function of initial velocity and acceleration.

Understanding these principles is key in predicting future positions and verifying that the equations hold true under given conditions.

Consistent Acceleration

Consistent acceleration means the acceleration (\( a \)) is constant throughout the motion. This is a vital assumption in the problem described. In physics, consistent acceleration makes calculations simpler because the same formulas can be applied over different intervals of time without variation.
In uniform acceleration, the particle's velocity increases at a constant rate, making its future positions predictable. This predictability is exploited in the problem to establish equations for distances covered at different times.

• Prediction: Easier due to consistent increase in speed.
• Simplicity: Uniform acceleration simplifies setting up equations and solving for unknowns.

Thus, understanding the importance of consistent acceleration is not only central to solving the problem but also critical in real-world applications where forces are uniform.