Two cars of same mass are moving with velocities `v_(1) and v_(2)` respectively. If they are stopped by supplying same breaking power in time `t_(1) and t_(2)` respectively then `(v_(1))/(v_(2))` is

Options

A. `(t_(1))/(t_(2))`
B. `(t_(1)^(2))/(t_(2)^(2))`
C. `m(t_(1))/(t_(2))`
D. `sqrt((t_(1))/(t_(2)))`

Answer

To solve the problem, we need to analyze the situation involving two cars with the same mass, moving at different velocities, and being stopped by the same braking power in different times.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Let the mass of both cars be \( m \).
- Let the velocities of the cars be \( v_1 \) and \( v_2 \).
- Let the times taken to stop be \( t_1 \) and \( t_2 \).
- The braking power is the same for both cars.

2. **Power and Force Relationship**:
- The power \( P \) can be expressed as:
\[ P = F \cdot v \]
- For car 1:
\[ P = F_1 \cdot v_1 \quad \text{(1)} \]
- For car 2:
\[ P = F_2 \cdot v_2 \quad \text{(2)} \]
- Since the braking power is the same for both cars, we have:
\[ F_1 \cdot v_1 = F_2 \cdot v_2 \quad \text{(3)} \]

3. **Acceleration Calculation**:
- The acceleration (or deceleration in this case) can be expressed using Newton's second law:
\[ F = m \cdot a \]
- Thus, the accelerations for the two cars are:
\[ a_1 = \frac{F_1}{m} \quad \text{and} \quad a_2 = \frac{F_2}{m} \]

4. **Using the Kinematic Equation**:
- For both cars, we can use the equation:
\[ v_f = v_i + a \cdot t \]
- For car 1 (final velocity \( v_f = 0 \)):
\[ 0 = v_1 - a_1 t_1 \quad \Rightarrow \quad a_1 = \frac{v_1}{t_1} \quad \text{(4)} \]
- For car 2:
\[ 0 = v_2 - a_2 t_2 \quad \Rightarrow \quad a_2 = \frac{v_2}{t_2} \quad \text{(5)} \]

5. **Relating the Forces**:
- From equations (4) and (5), we can express \( F_1 \) and \( F_2 \):
\[ F_1 = m \cdot a_1 = m \cdot \frac{v_1}{t_1} \quad \text{and} \quad F_2 = m \cdot a_2 = m \cdot \frac{v_2}{t_2} \]

6. **Substituting into Power Equation**:
- Substitute \( F_1 \) and \( F_2 \) into equation (3):
\[ \left(m \cdot \frac{v_1}{t_1}\right) v_1 = \left(m \cdot \frac{v_2}{t_2}\right) v_2 \]
- Canceling \( m \) and rearranging gives:
\[ \frac{v_1^2}{t_1} = \frac{v_2^2}{t_2} \]

7. **Finding the Ratio of Velocities**:
- Rearranging the above equation:
\[ \frac{v_1^2}{v_2^2} = \frac{t_1}{t_2} \]
- Taking the square root of both sides:
\[ \frac{v_1}{v_2} = \sqrt{\frac{t_1}{t_2}} \]

### Final Answer:
\[ \frac{v_1}{v_2} = \sqrt{\frac{t_1}{t_2}} \]