At the initial moment three points A, B and C are on a horizontal straight line at equal distances from one another. Point A begins to move vertically upward with a constant velocity v and point C vertically downward without any initial velocity but with a constant acceleration a. How should point B move vertically for all the three points to be constantly on one straight line. The points begin to move simultaneously.

Explanation

To solve the problem, we need to analyze the motion of the three points A, B, and C and determine how point B should move vertically to keep all three points in a straight line at all times.

### Step-by-Step Solution:

1. **Define the Initial Setup**:
- Let the distance between each point (A, B, and C) be \( D \).
- At \( t = 0 \), the positions are:
- A at (0, 0)
- B at (D, 0)
- C at (2D, 0)

2. **Motion of Point A**:
- Point A moves vertically upward with a constant velocity \( v \).
- The position of point A after time \( t \) is given by:
\[ y_A = vt \]

3. **Motion of Point C**:
- Point C moves vertically downward with an initial velocity of 0 and a constant acceleration \( a \).
- The position of point C after time \( t \) is given by:
\[ y_C = -\frac{1}{2} a t^2 \]

4. **Position of Point B**:
- Let the vertical displacement of point B after time \( t \) be \( y_B \).
- The position of point B after time \( t \) is given by:
\[ y_B = y_B(t) \]

5. **Condition for Collinearity**:
- For points A, B, and C to remain in a straight line, the slopes between A and B must equal the slopes between B and C at any time \( t \).
- This can be expressed using similar triangles:
\[ \frac{y_A - y_B}{D} = \frac{y_B - y_C}{D} \]
- Rearranging gives:
\[ y_A - y_B = y_B - y_C \]

6. **Substituting the Positions**:
- Substitute \( y_A \) and \( y_C \):
\[ vt - y_B = y_B + \frac{1}{2} a t^2 \]
- Rearranging leads to:
\[ vt = 2y_B + \frac{1}{2} a t^2 \]

7. **Solving for \( y_B \)**:
- Rearranging gives:
\[ 2y_B = vt - \frac{1}{2} a t^2 \]
- Thus:
\[ y_B = \frac{vt - \frac{1}{2} a t^2}{2} \]

8. **Finding the Velocity and Acceleration of Point B**:
- To find the velocity of point B, differentiate \( y_B \) with respect to \( t \):
\[ v_B = \frac{d}{dt}\left(\frac{vt - \frac{1}{2} a t^2}{2}\right) = \frac{1}{2}(v - at) \]
- To find the acceleration of point B, differentiate \( v_B \):
\[ a_B = \frac{d}{dt}\left(\frac{1}{2}(v - at)\right) = -\frac{a}{2} \]

9. **Conclusion**:
- Therefore, point B should move vertically with an initial velocity of \( \frac{v}{2} \) and an acceleration of \( -\frac{a}{2} \).

### Final Answer:
Point B should move vertically with an initial velocity of \( \frac{v}{2} \) and an acceleration of \( -\frac{a}{2} \).