a particle moving along a straight line with uniform acceleration has velocities `7 m//s` at A and `17 m//s` at C. B is the mid point of AC. Then :-

Options

A. The velocityt at B is `12 m//s`
B. The average velocity between A and B is `10 m//s`
C. The ratio of the time to go from A to B to that from B to C is `3:2`
D. The average velocity between B and C is `15 m//s`

Explanation

To solve the problem step by step, we will analyze the motion of the particle moving with uniform acceleration between points A, B, and C.

### Step 1: Understand the given information
- The velocity at point A (initial velocity, \( u \)) = 7 m/s
- The velocity at point C (final velocity, \( v \)) = 17 m/s
- Point B is the midpoint between A and C.

### Step 2: Use the equation of motion
We will use the equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity
- \( u \) = initial velocity
- \( a \) = acceleration
- \( s \) = displacement

### Step 3: Calculate the total displacement \( s \)
From point A to point C, the displacement is \( s \). Therefore, we can write:
\[ 17^2 = 7^2 + 2a s \]
Calculating \( 17^2 \) and \( 7^2 \):
\[ 289 = 49 + 2as \]
Rearranging gives:
\[ 2as = 289 - 49 = 240 \]
Thus, we have:
\[ as = 120 \quad \text{(Equation 1)} \]

### Step 4: Calculate the velocity at point B
Let the velocity at point B be \( v_B \). The displacement from A to B is \( \frac{s}{2} \). Using the same equation of motion for the segment AB:
\[ v_B^2 = u^2 + 2a\left(\frac{s}{2}\right) \]
Substituting the values:
\[ v_B^2 = 7^2 + 2a\left(\frac{s}{2}\right) \]
This simplifies to:
\[ v_B^2 = 49 + as \]
Now substituting \( as \) from Equation 1:
\[ v_B^2 = 49 + 120 = 169 \]
Taking the square root gives:
\[ v_B = 13 \text{ m/s} \]

### Step 5: Calculate the average velocity between A and B
The average velocity \( V_{avg} \) between points A and B is given by:
\[ V_{avg} = \frac{u + v_B}{2} = \frac{7 + 13}{2} = \frac{20}{2} = 10 \text{ m/s} \]

### Step 6: Calculate the ratio of time taken from A to B and B to C
Let \( T_1 \) be the time taken from A to B and \( T_2 \) be the time taken from B to C.

Using the formula for time:
\[ T_1 = \frac{v_B - u}{a} \quad \text{and} \quad T_2 = \frac{v_C - v_B}{a} \]

Substituting the values:
\[ T_1 = \frac{13 - 7}{a} = \frac{6}{a} \]
\[ T_2 = \frac{17 - 13}{a} = \frac{4}{a} \]
Now, the ratio \( \frac{T_1}{T_2} \) is:
\[ \frac{T_1}{T_2} = \frac{\frac{6}{a}}{\frac{4}{a}} = \frac{6}{4} = \frac{3}{2} \]

### Step 7: Calculate the average velocity between B and C
The average velocity \( V_{avg} \) between points B and C is given by:
\[ V_{avg} = \frac{v_B + v_C}{2} = \frac{13 + 17}{2} = \frac{30}{2} = 15 \text{ m/s} \]

### Summary of Results
- Velocity at point B: \( 13 \text{ m/s} \)
- Average velocity between A and B: \( 10 \text{ m/s} \)
- Ratio of time taken from A to B and B to C: \( 3:2 \)
- Average velocity between B and C: \( 15 \text{ m/s} \)