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 lt t lt 1)` . If `alpha ` denotes the instantaneous accelartion of the particle , then :

Options

A. A. `alpha` cannot remain positive for all t in the interval
B. B. `|alpha|` cannot exceed 2 at any point in its path.
C. C. `|alpha|` must be `gt`=`4` at some point or points in its path.
D. D. `alpha` must change sign during the motion, but no other assertion can be made
with the information given.

Explanation

To solve the problem, we need to analyze the motion of a particle that starts from rest, moves along the x-axis, and comes to rest again after a certain time. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the Initial and Final Conditions
- The particle starts from rest at \( t = 0 \) and \( x = 0 \). This means:
- Initial velocity \( u = 0 \)
- Initial position \( x_0 = 0 \)

- The particle comes to rest at \( t = 1 \) and \( x = 1 \). This means:
- Final velocity \( v = 0 \)
- Final position \( x_f = 1 \)

### Step 2: Analyze the Motion
Since the particle starts and ends at rest, it must accelerate at some point and then decelerate to come to rest. This implies that the acceleration cannot be positive throughout the entire motion.

### Step 3: Evaluate the Acceleration
- Let \( \alpha(t) \) denote the instantaneous acceleration of the particle.
- Since the particle starts from rest and comes to rest again, the acceleration must change sign during the motion:
- Initially, the particle must accelerate (positive acceleration) to gain speed.
- After reaching a certain speed, it must decelerate (negative acceleration) to come to rest.

### Step 4: Check the Options
1. **Option A**: "Alpha cannot remain positive for all \( t \) in the interval \( (0, 1) \)."
- This statement is true because the particle must decelerate to come to rest at \( t = 1 \). Therefore, option A is correct.

2. **Option B**: "Alpha cannot exceed 2 at any point of its path."
- To evaluate this, we can use the kinematic equation \( x = ut + \frac{1}{2} \alpha t^2 \). For the entire motion from \( t = 0 \) to \( t = 1 \):
\[ 1 = 0 \cdot 1 + \frac{1}{2} \alpha (1)^2 \implies \alpha = 2 \text{ m/s}^2 \]
- However, since the particle can accelerate more than this value before decelerating, option B is incorrect.

3. **Option C**: "Mod alpha must be greater than 4 at some point of its path."
- We can analyze the motion further, but since we found that the maximum acceleration can be 2 m/s², this statement is not necessarily true. Therefore, option C is incorrect.

4. **Option D**: "Alpha must change sign during the motion."
- This statement is true because the particle must accelerate and then decelerate, indicating that the acceleration must change from positive to negative. Therefore, option D is correct.

### Conclusion
The correct options are A and D.