A car starts moving on a straight road. It makes the entire journey in three parts. The total time of journey is 25 s. For first part it accelerates at a constant rate `5m//s^(2)`. For second part it moves with a constant speed. For the third part it retards at constant rate `5m//s^(2)` to come to rest.
If the average speed for entire journey is 20 m/s, time period of first part of journey is

Options

A. 5 s
B. 20 s
C. 15 s
D. 10 s

Answer

To solve the problem step-by-step, we will break down the journey of the car into three parts and use the given information to find the time period of the first part of the journey.

### Step 1: Understand the Problem
The car's journey consists of three parts:
1. Accelerating from rest with an acceleration of \(5 \, \text{m/s}^2\).
2. Moving with a constant speed.
3. Decelerating to come to rest with a deceleration of \(5 \, \text{m/s}^2\).

The total time for the journey is \(25 \, \text{s}\), and the average speed for the entire journey is \(20 \, \text{m/s}\).

### Step 2: Calculate Total Distance
The average speed is given by the formula:
\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \]
From this, we can calculate the total distance \(S\):
\[ S = \text{Average Speed} \times \text{Total Time} = 20 \, \text{m/s} \times 25 \, \text{s} = 500 \, \text{m} \]
This gives us our first equation:
\[ S = S_1 + S_2 + S_3 = 500 \, \text{m} \quad \text{(Equation 1)} \]

### Step 3: Set Up Time Relationships
Let:
- \(t_1\) = time for the first part (acceleration)
- \(t_2\) = time for the second part (constant speed)
- \(t_3\) = time for the third part (deceleration)

From the problem, we know:
\[ t_1 + t_2 + t_3 = 25 \, \text{s} \quad \text{(Equation 2)} \]

### Step 4: Analyze Each Part of the Journey
1. **First Part (Acceleration)**:
- Initial velocity \(u = 0\)
- Acceleration \(a = 5 \, \text{m/s}^2\)
- Final velocity \(v = u + at = 0 + 5t_1 = 5t_1\)
- Distance covered \(S_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 5 \times t_1^2 = \frac{5}{2} t_1^2\)

2. **Second Part (Constant Speed)**:
- Speed during this part is \(v = 5t_1\)
- Distance covered \(S_2 = v \cdot t_2 = (5t_1) \cdot t_2 = 5t_1 t_2\)

3. **Third Part (Deceleration)**:
- Initial velocity \(u = 5t_1\)
- Final velocity \(v = 0\)
- Deceleration \(a = -5 \, \text{m/s}^2\)
- Using \(v^2 = u^2 + 2as\):
\[ 0 = (5t_1)^2 - 2 \cdot 5 \cdot S_3 \implies S_3 = \frac{(5t_1)^2}{10} = \frac{5}{2} t_1^2 \]

### Step 5: Combine Distances
Now we can combine the distances:
\[ S_1 + S_2 + S_3 = \frac{5}{2} t_1^2 + 5t_1 t_2 + \frac{5}{2} t_1^2 = 500 \]
This simplifies to:
\[ 5t_1^2 + 5t_1 t_2 = 500 \]
Dividing through by \(5\):
\[ t_1^2 + t_1 t_2 = 100 \quad \text{(Equation 3)} \]

### Step 6: Substitute \(t_2\)
From Equation 2, we can express \(t_2\):
\[ t_2 = 25 - t_1 - t_3 \]
Since \(t_1 = t_3\) (from the symmetry of acceleration and deceleration):
\[ t_2 = 25 - 2t_1 \]

### Step 7: Substitute into Equation 3
Substituting \(t_2\) into Equation 3:
\[ t_1^2 + t_1(25 - 2t_1) = 100 \]
This expands to:
\[ t_1^2 + 25t_1 - 2t_1^2 = 100 \]
Combining like terms:
\[ -t_1^2 + 25t_1 - 100 = 0 \]
Multiplying through by \(-1\):
\[ t_1^2 - 25t_1 + 100 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ t_1 = \frac{25 \pm \sqrt{25^2 - 4 \cdot 1 \cdot 100}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 25^2 - 400 = 625 - 400 = 225 \]
Thus:
\[ t_1 = \frac{25 \pm 15}{2} \]
Calculating the two possible values:
1. \(t_1 = \frac{40}{2} = 20 \, \text{s}\)
2. \(t_1 = \frac{10}{2} = 5 \, \text{s}\)

Since \(t_1 + t_2 + t_3 = 25\) and \(t_1\) cannot be \(20\) (as \(t_2\) and \(t_3\) would be negative), we conclude:
\[ t_1 = 5 \, \text{s} \]

### Final Answer:
The time period of the first part of the journey is \(5 \, \text{s}\).