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

Answer

To solve the problem step by step, let's analyze the motion of the particle moving along a straight line with uniform acceleration.

### Given Data:
- Velocity at point A, \( V_A = 7 \, \text{m/s} \)
- Velocity at point C, \( V_C = 17 \, \text{m/s} \)
- B is the midpoint of AC.

### Step 1: Calculate the Velocity at Point B
Using the equation of motion for uniform acceleration, we know that:
\[ V_B^2 - V_A^2 = 2a \cdot s_{AB} \]
Where:
- \( V_B \) is the velocity at point B,
- \( s_{AB} \) is the distance from A to B,
- \( a \) is the acceleration.

Since B is the midpoint of AC, we can denote the total distance \( AC = S \), hence:
\[ s_{AB} = \frac{S}{2} \]
And the distance from B to C is also:
\[ s_{BC} = \frac{S}{2} \]

Using the equation for the entire distance from A to C:
\[ V_C^2 - V_A^2 = 2a \cdot S \]

Now we can express \( V_B^2 \) in terms of \( V_A \) and \( V_C \):
\[ V_B^2 - V_A^2 = 2a \cdot \frac{S}{2} \implies V_B^2 - V_A^2 = a \cdot S \]
We can substitute the expression for \( V_C \):
\[ V_B^2 - V_A^2 = \frac{1}{2}(V_C^2 - V_A^2) \]

Rearranging gives:
\[ V_B^2 = \frac{1}{2} V_C^2 + \frac{1}{2} V_A^2 \]

Substituting the known values:
\[ V_B^2 = \frac{1}{2} (17^2) + \frac{1}{2} (7^2) \]

Calculating:
\[ V_B^2 = \frac{1}{2} (289) + \frac{1}{2} (49) = \frac{289 + 49}{2} = \frac{338}{2} = 169 \]
\[ V_B = \sqrt{169} = 13 \, \text{m/s} \]

### Step 2: Average Velocity from A to B
The average velocity from A to B is given by:
\[ V_{\text{avg}} = \frac{V_A + V_B}{2} \]
Substituting the values:
\[ V_{\text{avg}} = \frac{7 + 13}{2} = \frac{20}{2} = 10 \, \text{m/s} \]

### Step 3: Time Ratio 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 \( V = u + at \):
1. From A to B:
\[ V_B = V_A + a t_1 \implies t_1 = \frac{V_B - V_A}{a} \]
2. From B to C:
\[ V_C = V_B + a t_2 \implies t_2 = \frac{V_C - V_B}{a} \]

The ratio of times is:
\[ \frac{t_1}{t_2} = \frac{V_B - V_A}{V_C - V_B} \]
Substituting the values:
\[ \frac{t_1}{t_2} = \frac{13 - 7}{17 - 13} = \frac{6}{4} = \frac{3}{2} \]

### Step 4: Average Velocity from B to C
The average velocity from B to C is given by:
\[ V_{\text{avg}} = \frac{V_B + V_C}{2} \]
Substituting the values:
\[ V_{\text{avg}} = \frac{13 + 17}{2} = \frac{30}{2} = 15 \, \text{m/s} \]

### Summary of Results:
1. Velocity at B, \( V_B = 13 \, \text{m/s} \)
2. Average velocity from A to B = \( 10 \, \text{m/s} \)
3. Ratio of time \( t_1 : t_2 = 3 : 2 \)
4. Average velocity from B to C = \( 15 \, \text{m/s} \)