Two bodies move in a straight line towards each other at initial velocities `v_(1)` and `v_(2)` and with constant acceleration `a_(1)` and `a_(2)` directed against the corresponding velocities at the initial instant. What must be the maximum initial separation between the bodies for which they meet during the motion ?

Explanation

To solve the problem of finding the maximum initial separation between two bodies moving towards each other with given initial velocities and accelerations, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- Let the two bodies be A and B.
- Body A has an initial velocity \( v_1 \) and acceleration \( a_1 \) (acting against its motion).
- Body B has an initial velocity \( v_2 \) and acceleration \( a_2 \) (acting against its motion).
- They are moving towards each other.

2. **Distance Covered by Each Body**:
- The distance covered by body A in time \( t \) can be expressed using the equation of motion:
\[ x_1 = v_1 t - \frac{1}{2} a_1 t^2 \]
- The distance covered by body B in time \( t \) is:
\[ x_2 = v_2 t - \frac{1}{2} a_2 t^2 \]

3. **Total Distance Between the Bodies**:
- The total distance \( x \) between the two bodies when they meet is the sum of the distances covered by both bodies:
\[ x = x_1 + x_2 \]
- Substituting the expressions for \( x_1 \) and \( x_2 \):
\[ x = \left(v_1 t - \frac{1}{2} a_1 t^2\right) + \left(v_2 t - \frac{1}{2} a_2 t^2\right) \]
- Simplifying this gives:
\[ x = (v_1 + v_2)t - \frac{1}{2}(a_1 + a_2)t^2 \]

4. **Condition for Meeting**:
- For the bodies to meet, the total distance \( x \) must be equal to the initial separation \( S \):
\[ S = (v_1 + v_2)t - \frac{1}{2}(a_1 + a_2)t^2 \]

5. **Finding the Time of Meeting**:
- To find the time \( t \) at which they meet, we can use the quadratic formula. Rearranging the equation gives:
\[ \frac{1}{2}(a_1 + a_2)t^2 - (v_1 + v_2)t + S = 0 \]
- This is a standard quadratic equation in the form \( At^2 + Bt + C = 0 \) where:
- \( A = \frac{1}{2}(a_1 + a_2) \)
- \( B = -(v_1 + v_2) \)
- \( C = S \)

6. **Discriminant Condition**:
- For the bodies to meet, the discriminant of this quadratic equation must be non-negative:
\[ D = B^2 - 4AC \geq 0 \]
- Substituting for \( A \), \( B \), and \( C \):
\[ (v_1 + v_2)^2 - 2(a_1 + a_2)S \geq 0 \]
- Rearranging gives:
\[ S \leq \frac{(v_1 + v_2)^2}{2(a_1 + a_2)} \]

7. **Conclusion**:
- The maximum initial separation \( S_{max} \) for which the two bodies meet is:
\[ S_{max} = \frac{(v_1 + v_2)^2}{2(a_1 + a_2)} \]