Two trains A and B 100 km apart are travelling to each other with starting speeds of `50 km//hr` for both. The train A accelerating at `18 km // hr^(2)` and B decelerating at `18 km//hr^(2)` . The distance where the engines cross each other from the initial position of A is

Options

A. 50 km
B. 68 km
C. 32 km
D. 59 km

Answer

To solve the problem, we need to determine the distance from the initial position of Train A where the two trains cross each other. Here’s the step-by-step solution:

### Step 1: Define the initial conditions
- Distance between the two trains (D) = 100 km
- Initial speed of Train A (u_A) = 50 km/hr
- Initial speed of Train B (u_B) = 50 km/hr
- Acceleration of Train A (a_A) = 18 km/hr²
- Deceleration of Train B (a_B) = -18 km/hr² (since it's decelerating)

### Step 2: Calculate the relative speed and acceleration
Since the trains are moving towards each other, we can consider their speeds and accelerations relative to each other:
- Relative initial speed (u_relative) = u_A + |u_B| = 50 + 50 = 100 km/hr
- Relative acceleration (a_relative) = a_A + |a_B| = 18 + 18 = 36 km/hr²

### Step 3: Use the equation of motion to find the time until they meet
We can use the equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
Where:
- S = distance covered (100 km)
- u = initial relative speed (100 km/hr)
- a = relative acceleration (36 km/hr²)

Setting up the equation:
\[ 100 = 100t + \frac{1}{2}(36)t^2 \]

### Step 4: Rearranging the equation
This simplifies to:
\[ 100 = 100t + 18t^2 \]
Rearranging gives:
\[ 18t^2 + 100t - 100 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where:
- a = 18
- b = 100
- c = -100

Calculating the discriminant:
\[ b^2 - 4ac = 100^2 - 4 \times 18 \times (-100) = 10000 + 7200 = 17200 \]

Now substituting into the quadratic formula:
\[ t = \frac{-100 \pm \sqrt{17200}}{2 \times 18} \]
Calculating \( \sqrt{17200} \approx 131.07 \):
\[ t = \frac{-100 \pm 131.07}{36} \]

Taking the positive root (since time cannot be negative):
\[ t = \frac{31.07}{36} \approx 0.86 \text{ hours} \]

### Step 6: Calculate the distance covered by Train A
Now we can find the distance covered by Train A in this time:
\[ \text{Distance}_A = u_A \cdot t + \frac{1}{2} a_A \cdot t^2 \]
Substituting the values:
\[ \text{Distance}_A = 50 \cdot 0.86 + \frac{1}{2} \cdot 18 \cdot (0.86)^2 \]
Calculating:
\[ \text{Distance}_A = 43 + \frac{1}{2} \cdot 18 \cdot 0.7396 \]
\[ \text{Distance}_A = 43 + 6.66 \approx 49.66 \text{ km} \]

### Conclusion
The distance from the initial position of Train A where the engines cross each other is approximately **49.66 km**.

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