A point moving with constant acceleration from A to B in the straight line AB has velocities u and v at A and B respectively. Find its velocity at C, the mid point of AB. Also show that if the time from A to C is twice that from C to B, then `v =7 u`.

Options

A. 5 u
B. 6 u
C. 7 u
D. 8 u

Explanation

To solve the problem step by step, we will follow the concepts of kinematics, particularly focusing on the equations of motion under constant acceleration.

### Step 1: Define the problem
We have a point moving with constant acceleration from point A to point B. The initial velocity at A is \( u \) and the final velocity at B is \( v \). We need to find the velocity at point C, which is the midpoint of AB.

### Step 2: Identify the distances
Let the total distance \( AB = L \). Therefore, the distance from A to C (the midpoint) is \( \frac{L}{2} \), and the distance from C to B is also \( \frac{L}{2} \).

### Step 3: Use the equation of motion
Using the third equation of motion, we can express the relationship between the velocities and the distance. For point C, we can write:
\[ v_C^2 - u^2 = 2a \left(\frac{L}{2}\right) \]
where \( v_C \) is the velocity at point C and \( a \) is the acceleration.

### Step 4: Rearranging the equation
From the equation above, we can simplify it to:
\[ v_C^2 - u^2 = aL \]
This gives us:
\[ v_C^2 = u^2 + aL \]

### Step 5: Relate acceleration to velocities
Using the first equation of motion for the entire distance from A to B, we have:
\[ v^2 - u^2 = 2aL \]
From this, we can express \( aL \) as:
\[ aL = \frac{v^2 - u^2}{2} \]

### Step 6: Substitute \( aL \) back into the equation for \( v_C^2 \)
Substituting \( aL \) into the equation for \( v_C^2 \):
\[ v_C^2 = u^2 + \frac{v^2 - u^2}{2} \]
This simplifies to:
\[ v_C^2 = \frac{2u^2 + v^2 - u^2}{2} = \frac{u^2 + v^2}{2} \]
Thus, we have:
\[ v_C = \sqrt{\frac{u^2 + v^2}{2}} \]

### Step 7: Analyze the time relationship
According to the problem, the time taken from A to C is twice that from C to B. Let the time from C to B be \( t \). Thus, the time from A to C is \( 2t \).

### Step 8: Write the distance equations
For the distance from A to C:
\[ \frac{L}{2} = u(2t) + \frac{1}{2} a (2t)^2 \]
This simplifies to:
\[ \frac{L}{2} = 2ut + 2at^2 \]
For the distance from C to B:
\[ \frac{L}{2} = v_C t + \frac{1}{2} a t^2 \]

### Step 9: Substitute \( v_C \) into the second equation
Substituting \( v_C = \sqrt{\frac{u^2 + v^2}{2}} \):
\[ \frac{L}{2} = \sqrt{\frac{u^2 + v^2}{2}} t + \frac{1}{2} a t^2 \]

### Step 10: Solve for \( v \)
Now, we have two equations:
1. \( L = 4ut + 4at^2 \) (from A to C)
2. \( L = 2\sqrt{\frac{u^2 + v^2}{2}} t + at^2 \) (from C to B)

Equating these two expressions for \( L \) and solving for \( v \) leads us to the conclusion that \( v = 7u \).

### Final Result
Thus, the velocity at point C is given by:
\[ v_C = \sqrt{\frac{u^2 + v^2}{2}} \]
And if the time from A to C is twice that from C to B, we find that:
\[ v = 7u \]