A particle moves with uniform acceleration along a straight line `AB`. Its velocities at `A and B` are `2 m//s and 14 m//s` respectively. `M` is the mid-point of `AB`. The particle takes `t_1` seconds to go from `A` to `M` and `t_2` seconds to go from `M` to `B`. Then `t_2 : t_1` is.

Options

A. `1 : 1`
B. `2 : 1`
C. `1 : 2`
D. `3 : 1`

Explanation

To solve the problem, we need to find the ratio \( t_2 : t_1 \) for a particle moving with uniform acceleration from point A to B, where the velocities at A and B are given as \( 2 \, \text{m/s} \) and \( 14 \, \text{m/s} \) respectively.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Initial velocity at A, \( u = 2 \, \text{m/s} \)
- Final velocity at B, \( v = 14 \, \text{m/s} \)

2. **Use the Equation of Motion**:
We can use the equation of motion:
\[ v^2 = u^2 + 2as \]
where \( s \) is the total distance from A to B, and \( a \) is the acceleration.

3. **Calculate the Total Distance**:
Let the distance from A to B be \( s \). The midpoint M divides this distance into two equal parts, so:
\[ s_A = s_B = \frac{s}{2} \]

4. **Apply the Equation of Motion for A to M**:
For the first half (A to M):
\[ v_M^2 = u^2 + 2a\left(\frac{s}{2}\right) \]
Let \( v_M \) be the velocity at M. Thus:
\[ v_M^2 = 2^2 + 2a\left(\frac{s}{2}\right) \quad \text{(1)} \]

5. **Apply the Equation of Motion for M to B**:
For the second half (M to B):
\[ v^2 = v_M^2 + 2a\left(\frac{s}{2}\right) \]
Thus:
\[ 14^2 = v_M^2 + 2a\left(\frac{s}{2}\right) \quad \text{(2)} \]

6. **Substituting the Equations**:
From equation (1):
\[ v_M^2 = 4 + as \]
Substitute \( v_M^2 \) in equation (2):
\[ 196 = (4 + as) + as \]
Simplifying gives:
\[ 196 = 4 + 2as \]
\[ 192 = 2as \quad \Rightarrow \quad as = 96 \quad \text{(3)} \]

7. **Finding \( t_1 \) and \( t_2 \)**:
Now we can find \( t_1 \) and \( t_2 \):
- For \( t_1 \) (A to M):
\[ v_M = u + at_1 \quad \Rightarrow \quad v_M = 2 + at_1 \]
Rearranging gives:
\[ t_1 = \frac{v_M - 2}{a} \]

- For \( t_2 \) (M to B):
\[ 14 = v_M + at_2 \quad \Rightarrow \quad t_2 = \frac{14 - v_M}{a} \]

8. **Finding the Ratio \( t_2 : t_1 \)**:
Now, substituting \( v_M \):
From (3), we know \( as = 96 \), thus:
\[ a = \frac{96}{s} \]
Substitute \( a \) in \( t_1 \) and \( t_2 \):
\[ t_1 = \frac{v_M - 2}{\frac{96}{s}} = \frac{s(v_M - 2)}{96} \]
\[ t_2 = \frac{14 - v_M}{\frac{96}{s}} = \frac{s(14 - v_M)}{96} \]

Now, the ratio:
\[ \frac{t_2}{t_1} = \frac{s(14 - v_M)}{s(v_M - 2)} = \frac{14 - v_M}{v_M - 2} \]

9. **Finding \( v_M \)**:
From the earlier calculations, we can find \( v_M \):
Since \( as = 96 \) and \( v_M^2 = 4 + as \):
\[ v_M^2 = 4 + 96 = 100 \quad \Rightarrow \quad v_M = 10 \, \text{m/s} \]

10. **Final Calculation of the Ratio**:
Substitute \( v_M = 10 \):
\[ \frac{t_2}{t_1} = \frac{14 - 10}{10 - 2} = \frac{4}{8} = \frac{1}{2} \]

### Conclusion:
The ratio \( t_2 : t_1 \) is \( 1 : 2 \).