A particle starts from rest, accelerates uniformly for 3 seconds and then decelerates uniformly for 3 seconds and comes to rest. Which one of the following displacement (x)-time (t) graphs represents the motion of the particle?

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the motion of the particle step by step.

### Step 1: Understanding the Motion
The particle starts from rest, accelerates uniformly for 3 seconds, and then decelerates uniformly for another 3 seconds until it comes to rest.

### Step 2: Acceleration Phase
During the first 3 seconds, the particle accelerates from rest.
- Initial velocity (u) = 0 m/s
- Time (t) = 3 seconds
- Since the particle accelerates uniformly, we can use the equations of motion.
- The displacement (s) during this phase can be calculated using the formula:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \(u = 0\), so:
\[ s = 0 + \frac{1}{2} a (3)^2 = \frac{9a}{2} \]
- The final velocity (v) at the end of this phase can be calculated using:
\[ v = u + at = 0 + 3a = 3a \]

### Step 3: Deceleration Phase
In the next 3 seconds, the particle decelerates uniformly to come to rest.
- Initial velocity (u) = 3a (from the end of the first phase)
- Final velocity (v) = 0 m/s
- Time (t) = 3 seconds
- The deceleration (a') can be calculated using:
\[ v = u + a't \implies 0 = 3a + a' (3) \implies a' = -a \]
- The displacement (s') during this phase can be calculated as:
\[ s' = ut + \frac{1}{2} a' t^2 = 3a(3) + \frac{1}{2} (-a)(3^2) = 9a - \frac{9a}{2} = \frac{9a}{2} \]

### Step 4: Total Displacement
The total displacement during the entire motion is:
\[ s_{total} = s + s' = \frac{9a}{2} + \frac{9a}{2} = 9a \]
The particle starts from rest, accelerates for 3 seconds, reaches a maximum displacement, and then decelerates for another 3 seconds, returning to rest.

### Step 5: Graph Analysis
Now, we need to analyze the displacement-time (x-t) graph:
- For the first 3 seconds, the graph will be a curve that increases (since the particle is accelerating).
- For the next 3 seconds, the graph will be a curve that decreases (since the particle is decelerating).
- The graph should start at the origin (0,0) and return to the x-axis at the end of 6 seconds.

### Conclusion
The correct displacement-time graph will show a curve that rises for the first 3 seconds and then falls for the next 3 seconds, forming a parabolic shape.