A car starts moving along a line, first with acceleration `a=5m//s^(2)` starting from rest then uniformly and finally decelerating at the same rate till it comes to rest. The total time of motion is `25s`. The average speed during the time is `20m//s`. The particle moves uniformly for `(2.5x)` second. Find the value of `x`.

Answer

To solve the problem step by step, we will analyze the motion of the car in three phases: acceleration, uniform motion, and deceleration.

### Step 1: Define the phases of motion
1. **Acceleration Phase (A to C)**: The car accelerates from rest with an acceleration of \( a = 5 \, \text{m/s}^2 \) for a time \( t \).
2. **Uniform Motion Phase (C to D)**: The car moves uniformly for a time \( T \).
3. **Deceleration Phase (D to B)**: The car decelerates at the same rate of \( -5 \, \text{m/s}^2 \) for the same time \( t \).

### Step 2: Express total time
The total time of motion is given as \( 25 \, \text{s} \). Therefore, we can express the total time as:
\[ T + 2t = 25 \]
where \( T \) is the time spent in uniform motion.

### Step 3: Average speed and total distance
The average speed is given as \( 20 \, \text{m/s} \). The formula for average speed is:
\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \]
Thus, we have:
\[ 20 = \frac{\text{Total Distance}}{25} \]
This implies:
\[ \text{Total Distance} = 20 \times 25 = 500 \, \text{m} \]

### Step 4: Calculate distances for each phase
1. **Distance from A to C** (accelerating):
\[ s_{AC} = \frac{1}{2} a t^2 = \frac{1}{2} \times 5 \times t^2 = \frac{5}{2} t^2 \]

2. **Distance from C to D** (uniform motion):
The velocity at the end of the acceleration phase (at point C) is:
\[ v_C = a \cdot t = 5t \]
Therefore, the distance during uniform motion is:
\[ s_{CD} = v_C \cdot T = 5t \cdot T \]

3. **Distance from D to B** (decelerating):
The distance during deceleration is:
\[ s_{DB} = v_C \cdot t + \frac{1}{2} (-a) t^2 = 5t \cdot t - \frac{1}{2} \cdot 5 \cdot t^2 = 5t^2 - \frac{5}{2} t^2 = \frac{5}{2} t^2 \]

### Step 5: Total distance equation
Now, we can express the total distance as:
\[ \text{Total Distance} = s_{AC} + s_{CD} + s_{DB} \]
Substituting the distances:
\[ 500 = \frac{5}{2} t^2 + 5tT + \frac{5}{2} t^2 \]
This simplifies to:
\[ 500 = 5t^2 + 5tT \]
Dividing the entire equation by 5:
\[ 100 = t^2 + tT \]

### Step 6: Substitute \( T \) in terms of \( t \)
From the total time equation \( T = 25 - 2t \), we substitute \( T \):
\[ 100 = t^2 + t(25 - 2t) \]
Expanding this gives:
\[ 100 = t^2 + 25t - 2t^2 \]
Rearranging leads to:
\[ 0 = -t^2 + 25t - 100 \]
Multiplying through by -1:
\[ t^2 - 25t + 100 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{25 \pm \sqrt{625 - 400}}{2} \]
\[ t = \frac{25 \pm \sqrt{225}}{2} \]
\[ t = \frac{25 \pm 15}{2} \]
Thus, we have:
\[ t = 20 \quad \text{or} \quad t = 5 \]
Since \( t \) cannot be 20 (as it would not allow for uniform motion), we take \( t = 5 \).

### Step 8: Calculate \( T \) and \( x \)
Now substituting \( t \) back into the equation for \( T \):
\[ T = 25 - 2 \cdot 5 = 15 \, \text{s} \]
Given that the particle moves uniformly for \( 2.5x \) seconds:
\[ 2.5x = T = 15 \]
Thus:
\[ x = \frac{15}{2.5} = 6 \]

### Final Answer
The value of \( x \) is:
\[ \boxed{6} \]