A particle of mass m moves on the x-axis as follows: it starts from rest at t = 0 from the point x = 0, and comes to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate times ( 0

If α denotes the magnitude of instantaneous acceleration of the particle, then which of the following is/are true?
(a) α cannot remain positive throughout
(b) α cannot exceed 2 at any point in its path
(c) α must be equal to 4 at some point or points in its path
(d) α must change sign during the motion, but no other assertion can be made with the information given

Obviously the a part is correct because the the particle has to deaccelerate to come to a stop.
i think one can do this by shm also but please give me the solution with the help of kinematics also. I dont understand why the c part is correct.

To analyze the motion of the particle and determine the validity of the statements regarding its instantaneous acceleration, we can use basic kinematic principles. Let's break down the problem step by step.

Understanding the Motion

The particle starts from rest at \( t = 0 \) from the point \( x = 0 \) and comes to rest at \( t = 1 \) at the point \( x = 1 \). This means that the particle travels a distance of 1 unit in 1 second. Since it starts and ends at rest, we can infer that its velocity changes throughout the motion.

Analyzing the Kinematics

We can use the kinematic equations to understand the relationship between displacement, velocity, and acceleration. The average velocity \( v_{avg} \) can be calculated as:

• Average velocity \( v_{avg} = \frac{\Delta x}{\Delta t} = \frac{1 - 0}{1 - 0} = 1 \, \text{unit/s} \)

Since the particle starts and ends at rest, the initial velocity \( v_0 = 0 \) and the final velocity \( v_f = 0 \). The average velocity being 1 unit/s indicates that the particle must have had some positive velocity during its motion, which means it accelerated at some point and then decelerated to come to rest.

Acceleration Analysis

The instantaneous acceleration \( \alpha \) is defined as the rate of change of velocity. Since the particle starts from rest and comes to rest again, the acceleration must change throughout the motion. Let's evaluate the statements one by one:

Statement (a): α cannot remain positive throughout

This statement is true. Since the particle starts from rest, it must accelerate initially to gain speed, but to come to rest at \( t = 1 \), it must eventually decelerate. Therefore, the acceleration cannot remain positive throughout the motion.

Statement (b): α cannot exceed 2 at any point in its path

This statement is also true. We can use the kinematic equation that relates acceleration, initial velocity, final velocity, and displacement:

• Using \( v^2 = u^2 + 2a s \), where \( u = 0 \), \( v = 0 \), and \( s = 1 \):
• Setting \( 0 = 0 + 2a(1) \) gives \( a = 0 \). However, we need to consider the average acceleration over the entire motion.
• The average acceleration \( a_{avg} = \frac{\Delta v}{\Delta t} = \frac{0 - 0}{1} = 0 \). However, since the particle must accelerate and decelerate, the maximum instantaneous acceleration cannot exceed 2 units/s² to ensure it can come to rest at the end of the interval.

Statement (c): α must be equal to 4 at some point or points in its path

This statement is true as well. To see why, consider the average acceleration over the entire motion. The particle must accelerate to reach a maximum speed and then decelerate to come to rest. The maximum speed \( v_{max} \) can be calculated using the average velocity:

• Since the average velocity is 1 unit/s, the maximum speed must occur at some point before it starts decelerating. If we assume a triangular velocity-time graph, the area under the graph (which represents displacement) must equal 1.
• To achieve this, the maximum instantaneous acceleration could reach 4 units/s² at some point, allowing the particle to reach a sufficient speed before decelerating back to rest.

Statement (d): α must change sign during the motion, but no other assertion can be made with the information given

This statement is also true. As previously discussed, the particle must accelerate to gain speed and then decelerate to come to rest, indicating that the acceleration must change from positive to negative at some point during the motion.

Summary of Findings

In conclusion, all four statements can be validated based on the kinematic analysis of the particle's motion:

• (a) True: α cannot remain positive throughout.
• (b) True: α cannot exceed 2 at any point in its path.
• (c) True: α must be equal to 4 at some point or points in its path.
• (d) True: α must change sign during the motion.

Understanding these principles helps clarify the relationship between displacement, velocity, and acceleration in motion. If you have any further questions or need clarification on any point, feel free to ask!