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 determine how point B should move vertically so that points A, B, and C remain in a straight line at all times. Let's break down the solution step by step.

### Step 1: Understand the Initial Setup
We have three points A, B, and C on a horizontal line, equally spaced. Let's denote the distance between each point as \(d\). At time \(t = 0\):
- Point A is at position \(A(0, 0)\) and moves upward with a constant velocity \(v\).
- Point B is at position \(B(0, d)\) and will move vertically (we need to determine how).
- Point C is at position \(C(0, 2d)\) and moves downward with an initial velocity of 0 and a constant acceleration \(a\).

### Step 2: Write the Equations of Motion
For point A moving upward:
- The position of A at time \(t\) is given by:
\[ y_A = vt \]

For point C moving downward:
- The position of C at time \(t\) is given by:
\[ y_C = 2d - \frac{1}{2} a t^2 \]
(Note: The downward motion is considered negative.)

Let the position of point B at time \(t\) be:
\[ y_B = y_B(t) \]
We need to find the function \(y_B(t)\) such that points A, B, and C remain collinear.

### Step 3: Condition for Collinearity
For points A, B, and C to be collinear, the slopes between these points must be equal. This can be expressed as:
\[ \frac{y_A - y_B}{0 - 0} = \frac{y_B - y_C}{0 - 0} \]
This condition simplifies to:
\[ (y_A - y_B) = k(y_B - y_C) \]
where \(k\) is a proportionality constant that remains the same for the slopes.

### Step 4: Substitute the Positions
Substituting the equations of motion into the collinearity condition:
\[ (vt - y_B) = k(y_B - (2d - \frac{1}{2} a t^2)) \]

### Step 5: Solve for \(y_B\)
Rearranging gives:
\[ vt - y_B = k(y_B - 2d + \frac{1}{2} a t^2) \]
This can be solved for \(y_B\) to find how point B must move.

### Step 6: Equate and Simplify
By isolating \(y_B\) and simplifying, we can find the relationship between \(v\), \(a\), and the motion of point B. After some algebra, we find:
\[ y_B = \frac{v}{2}t - \frac{a}{2}t^2 + \text{(constant)} \]

### Step 7: Final Result
The final expression indicates that point B must move with an initial velocity of \(\frac{v}{2}\) and an acceleration of \(-\frac{a}{2}\) in the vertical direction.

### Summary
To keep points A, B, and C in a straight line:
- Point B should move vertically with an initial velocity of \(\frac{v}{2}\) and an acceleration of \(-\frac{a}{2}\).