A sprinter accelerates from rest to a top speed with an acceleration whose magnitude is `3.80 m//s^(2)`. After achieving top speed, he runs the remainder of the race without speeding up or slowing down. The total race is fifty meters long. If the total race is run in 7.88 s, how far does the run during the acceleration phase ?

Options

A. 6.85 m
B. 9.03 m
C. 7.62 m
D. 13.6 m

Answer

To solve the problem step by step, we will break it down into manageable parts.

### Given:
- Total distance of the race, \( S = 50 \, \text{m} \)
- Total time taken, \( T = 7.88 \, \text{s} \)
- Acceleration, \( a = 3.80 \, \text{m/s}^2 \)
- Initial velocity, \( u = 0 \, \text{m/s} \)

### Objective:
Find the distance covered during the acceleration phase, denoted as \( S_1 \).

### Step 1: Define the phases of motion
The motion can be divided into two phases:
1. **Acceleration Phase (from rest to top speed)**:
- Distance covered: \( S_1 \)
- Time taken: \( t_1 \)
- Final velocity at the end of this phase: \( v \)

2. **Constant Speed Phase (top speed to finish)**:
- Distance covered: \( S_2 \)
- Time taken: \( t_2 \)

### Step 2: Write the equations for each phase
For the acceleration phase, we can use the equation of motion:
\[ S_1 = ut + \frac{1}{2} a t_1^2 \]
Since \( u = 0 \):
\[ S_1 = \frac{1}{2} a t_1^2 = \frac{1}{2} \times 3.80 \times t_1^2 = 1.9 t_1^2 \tag{1} \]

For the constant speed phase, the distance covered can be expressed as:
\[ S_2 = v \cdot t_2 \]
Where \( v \) is the final velocity at the end of the acceleration phase. Using the first equation of motion:
\[ v = u + at_1 = 0 + 3.80 t_1 = 3.80 t_1 \tag{2} \]
Thus,
\[ S_2 = (3.80 t_1) t_2 \tag{3} \]

### Step 3: Relate the total distance and time
From the problem, we know:
\[ S_1 + S_2 = 50 \tag{4} \]
And,
\[ t_1 + t_2 = 7.88 \tag{5} \]

### Step 4: Substitute \( t_2 \) in terms of \( t_1 \)
From equation (5):
\[ t_2 = 7.88 - t_1 \tag{6} \]

### Step 5: Substitute \( S_2 \) and \( t_2 \) in equation (4)
Substituting equations (1), (3), and (6) into equation (4):
\[ 1.9 t_1^2 + (3.80 t_1)(7.88 - t_1) = 50 \]

### Step 6: Simplify and solve for \( t_1 \)
Expanding the equation:
\[ 1.9 t_1^2 + 3.80 \cdot 7.88 t_1 - 3.80 t_1^2 = 50 \]
Combining like terms:
\[ (1.9 - 3.80) t_1^2 + 29.944 t_1 - 50 = 0 \]
This simplifies to:
\[ -1.9 t_1^2 + 29.944 t_1 - 50 = 0 \]
Multiplying through by -1:
\[ 1.9 t_1^2 - 29.944 t_1 + 50 = 0 \]

### Step 7: Use the quadratic formula to solve for \( t_1 \)
Using the quadratic formula \( t_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where \( a = 1.9, b = -29.944, c = 50 \):
\[ t_1 = \frac{29.944 \pm \sqrt{(-29.944)^2 - 4 \cdot 1.9 \cdot 50}}{2 \cdot 1.9} \]
Calculating the discriminant:
\[ (-29.944)^2 - 4 \cdot 1.9 \cdot 50 = 896.0 - 380 = 516.0 \]
Now substituting back:
\[ t_1 = \frac{29.944 \pm \sqrt{516}}{3.8} \]
Calculating \( \sqrt{516} \approx 22.7 \):
\[ t_1 = \frac{29.944 \pm 22.7}{3.8} \]
Calculating the two possible values for \( t_1 \):
1. \( t_1 \approx \frac{52.644}{3.8} \approx 13.85 \) (not valid as it exceeds total time)
2. \( t_1 \approx \frac{7.244}{3.8} \approx 1.9 \, \text{s} \) (valid)

### Step 8: Calculate \( S_1 \)
Substituting \( t_1 \) back into equation (1):
\[ S_1 = 1.9 t_1^2 = 1.9 \times (1.9)^2 = 1.9 \times 3.61 \approx 6.86 \, \text{m} \]

### Final Answer:
The distance covered during the acceleration phase \( S_1 \approx 6.86 \, \text{m} \).

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