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.

Options

A. acceleration a/2 and initial velocity v/2 both upwards
B. retardation a/2 and initial velocity v/2 both upwards
C. zero acceleration and initial velocity v
D. acceleration a/2 and initial velocity v both downwards

Explanation

To solve the problem, we need to analyze the motion of points A, B, and C and ensure that they remain in a straight line as they move. Here’s a step-by-step solution:

### Step 1: Define the Initial Positions
Let the initial positions of points A, B, and C be:
- A at (0, 0)
- B at (D, 0)
- C at (2D, 0)

Here, D is the distance between each point.

### Step 2: Define the Motion of Points
- Point A moves vertically upward with a constant velocity \( v \). Therefore, its position after time \( t \) will be:
\[ A' = (0, vt) \]

- Point C moves vertically downward with an initial velocity of 0 and a constant acceleration \( a \). Its position after time \( t \) will be:
\[ C' = (2D, -\frac{1}{2} a t^2) \]

### Step 3: Determine the Position of Point B
Let point B move vertically with an unknown velocity \( u \) and acceleration \( b \). Its position after time \( t \) will be:
\[ B' = (D, ut + \frac{1}{2} b t^2) \]

### Step 4: Ensure Collinearity of Points A, B, and C
For points A, B, and C to be collinear, the slopes of the lines connecting these points must be equal. We can use the concept of similar triangles.

The slope of line segment \( A'C' \) is:
\[ \text{slope}_{A'C'} = \frac{C_y' - A_y'}{C_x' - A_x'} = \frac{-\frac{1}{2} a t^2 - vt}{2D - 0} = \frac{-\frac{1}{2} a t^2 - vt}{2D} \]

The slope of line segment \( A'B' \) is:
\[ \text{slope}_{A'B'} = \frac{B_y' - A_y'}{B_x' - A_x'} = \frac{ut + \frac{1}{2} b t^2 - vt}{D - 0} = \frac{ut + \frac{1}{2} b t^2 - vt}{D} \]

### Step 5: Set the Slopes Equal
For collinearity, set the two slopes equal:
\[ \frac{-\frac{1}{2} a t^2 - vt}{2D} = \frac{ut + \frac{1}{2} b t^2 - vt}{D} \]

### Step 6: Simplify the Equation
Cross-multiplying gives:
\[ -\frac{1}{2} a t^2 - vt = 2(ut + \frac{1}{2} b t^2 - vt) \]
Expanding and rearranging terms leads to:
\[ -\frac{1}{2} a t^2 - vt = 2ut + bt^2 - 2vt \]
\[ -\frac{1}{2} a t^2 + vt = 2ut + bt^2 \]

### Step 7: Solve for \( u \) and \( b \)
Rearranging gives:
\[ vt + \frac{1}{2} a t^2 = 2ut + bt^2 \]
This can be rearranged to express \( u \) and \( b \) in terms of \( v \) and \( a \):
\[ u = \frac{v}{2} \quad \text{and} \quad b = -\frac{a}{2} \]

### Conclusion
Thus, point B should move vertically with an initial velocity of \( \frac{v}{2} \) upwards and an acceleration of \( -\frac{a}{2} \) downwards.