A train starts from rest with constant acceleration `a=1 m//s^(2)` A passenger at a distance S from the train runs at this maximum velocity of `10 m//s` to catch the train at the same moment at which the train starts.
Find the speed of the train when the passenger catches it for the critical distance:

Options

A. `8 m//s`
B. `10 m//s`
C. `12 m//s`
D. `15 m//s`

Explanation

To solve the problem step-by-step, we will analyze the motion of both the train and the passenger, and find the speed of the train when the passenger catches it at the critical distance.

### Step 1: Understand the motion of the train
The train starts from rest with a constant acceleration \( a = 1 \, \text{m/s}^2 \). The initial velocity \( u \) of the train is \( 0 \, \text{m/s} \).

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Since \( u = 0 \):
\[ s = \frac{1}{2} a t^2 = \frac{1}{2} (1) t^2 = \frac{t^2}{2} \]

### Step 2: Understand the motion of the passenger
The passenger runs at a constant speed of \( v_p = 10 \, \text{m/s} \). The distance covered by the passenger when he catches the train is given by:
\[ d_p = v_p \cdot t = 10t \]

### Step 3: Set up the equation for the distances
The passenger starts at a distance \( S = 25.5 \, \text{m} \) behind the train. When the passenger catches the train, the total distance covered by the passenger will be equal to the distance covered by the train plus the initial distance \( S \):
\[ d_p = S + d_t \]
Substituting the expressions for \( d_p \) and \( d_t \):
\[ 10t = 25.5 + \frac{t^2}{2} \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ \frac{t^2}{2} - 10t + 25.5 = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ t^2 - 20t + 51 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -20, c = 51 \):
\[ t = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 51}}{2 \cdot 1} \]
\[ t = \frac{20 \pm \sqrt{400 - 204}}{2} \]
\[ t = \frac{20 \pm \sqrt{196}}{2} \]
\[ t = \frac{20 \pm 14}{2} \]
This gives two solutions:
\[ t = \frac{34}{2} = 17 \, \text{s} \quad \text{and} \quad t = \frac{6}{2} = 3 \, \text{s} \]

### Step 6: Determine the relevant time
The passenger catches the train first at \( t = 3 \, \text{s} \). The second time \( t = 17 \, \text{s} \) indicates that the passenger will eventually be caught by the train again after running ahead.

### Step 7: Calculate the speed of the train when the passenger catches it
Using the formula for final velocity:
\[ v = u + at \]
Substituting \( u = 0 \, \text{m/s} \), \( a = 1 \, \text{m/s}^2 \), and \( t = 10 \, \text{s} \) (since we are interested in the critical distance):
\[ v = 0 + 1 \cdot 10 = 10 \, \text{m/s} \]

### Final Answer
The speed of the train when the passenger catches it for the critical distance is \( 10 \, \text{m/s} \).

---