Two trains travelling on the same track are approaching each other with equal speed of ` 40 m//s`. The drivers of the trains beging to decelerate simultaneously when just ` 2.0 km apart. Assuming deceleration to be uniform and equal the value to the deceleration to barely avoid collision should be .

Options

A. (a) ` 0.8 m //s^2`
B. (b) ` 2. 1 m//d^2`
C. (c ) ` 11 .0 m//d^2`
D. (d) ` 11 .8 m//d^`

Explanation

To solve the problem of two trains approaching each other and needing to decelerate to avoid a collision, we can follow these steps:

### Step 1: Understand the given data
- Both trains are traveling towards each other with a speed of \(40 \, \text{m/s}\).
- The distance between the two trains when they start decelerating is \(2.0 \, \text{km}\) or \(2000 \, \text{m}\).
- We need to find the uniform deceleration required for both trains to avoid a collision.

### Step 2: Calculate the initial relative speed
Since both trains are moving towards each other, their relative speed is the sum of their individual speeds:
\[ \text{Relative speed} = 40 \, \text{m/s} + 40 \, \text{m/s} = 80 \, \text{m/s} \]

### Step 3: Use the equation of motion
We will use the equation of motion that relates initial velocity, final velocity, acceleration, and distance:
\[ v^2 = u^2 + 2as \]
Where:
- \(v\) = final velocity (which is \(0 \, \text{m/s}\) since they need to stop)
- \(u\) = initial velocity (which is \(80 \, \text{m/s}\))
- \(a\) = acceleration (which will be negative since it's deceleration)
- \(s\) = distance (which is \(2000 \, \text{m}\))

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
\[ 0 = (80)^2 + 2 \cdot a \cdot 2000 \]
This simplifies to:
\[ 0 = 6400 + 4000a \]

### Step 5: Solve for acceleration
Rearranging the equation to solve for \(a\):
\[ 4000a = -6400 \]
\[ a = -\frac{6400}{4000} = -1.6 \, \text{m/s}^2 \]

### Step 6: Determine the deceleration
Since we are looking for the magnitude of deceleration:
\[ \text{Deceleration} = 1.6 \, \text{m/s}^2 \]

### Conclusion
The required uniform deceleration for both trains to barely avoid a collision is \(1.6 \, \text{m/s}^2\).

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