In a car race car `A` takes `t_(0)` time less to finish than car `B` and pases the finishing point with a velocity `v_(0)` more than car `B`. The cars start from rest and travel with constant accelerations `a_(1)` and `a_(2)` . Then the ratio `(v_(0))/(t_(0))` is equal to

Options

A. `(a_(1)^(2))/(a_(2))`
B. `(a_(1)+a_(2))/2`
C. `sqrt(a_(1)a_(2))`
D. `(a_(2)^(2))/(a_(1))`

Answer

To solve the problem, we need to analyze the motion of both cars A and B using the equations of kinematics.

### Step-by-Step Solution:

1. **Define Variables:**
- Let \( t_1 \) be the time taken by car A to finish the race.
- Since car A takes \( t_0 \) time less than car B, the time taken by car B will be \( t_1 + t_0 \).
- Let \( v_1 \) be the final velocity of car B, then the final velocity of car A will be \( v_1 + v_0 \).

2. **Kinematic Equations:**
- For car A (starting from rest):
\[ v_1 + v_0 = a_1 t_1 \quad \text{(1)} \]
- For car B (starting from rest):
\[ v_1 = a_2 (t_1 + t_0) \quad \text{(2)} \]

3. **Express \( v_1 \) from Equation (2):**
- Rearranging Equation (2):
\[ v_1 = a_2 (t_1 + t_0) \]

4. **Substitute \( v_1 \) into Equation (1):**
- Substitute \( v_1 \) from Equation (2) into Equation (1):
\[ a_2 (t_1 + t_0) + v_0 = a_1 t_1 \]
- Rearranging gives:
\[ v_0 = a_1 t_1 - a_2 (t_1 + t_0) \]
- Simplifying:
\[ v_0 = (a_1 - a_2) t_1 - a_2 t_0 \quad \text{(3)} \]

5. **Distance Covered by Both Cars:**
- The distance covered by car A:
\[ S_A = \frac{1}{2} a_1 t_1^2 \]
- The distance covered by car B:
\[ S_B = \frac{1}{2} a_2 (t_1 + t_0)^2 \]
- Since both cars cover the same distance:
\[ \frac{1}{2} a_1 t_1^2 = \frac{1}{2} a_2 (t_1 + t_0)^2 \]
- Canceling \( \frac{1}{2} \) gives:
\[ a_1 t_1^2 = a_2 (t_1^2 + 2t_1 t_0 + t_0^2) \]

6. **Rearranging the Distance Equation:**
- Rearranging gives:
\[ a_1 t_1^2 = a_2 t_1^2 + 2a_2 t_1 t_0 + a_2 t_0^2 \]
- This can be rearranged to:
\[ (a_1 - a_2) t_1^2 - 2a_2 t_1 t_0 - a_2 t_0^2 = 0 \]

7. **Solving the Quadratic Equation:**
- This is a quadratic equation in \( t_1 \). Using the quadratic formula:
\[ t_1 = \frac{-(-2a_2 t_0) \pm \sqrt{(2a_2 t_0)^2 - 4(a_1 - a_2)(-a_2 t_0^2)}}{2(a_1 - a_2)} \]
- Simplifying gives:
\[ t_1 = \frac{2a_2 t_0 \pm \sqrt{4a_2^2 t_0^2 + 4a_2 t_0^2 (a_1 - a_2)}}{2(a_1 - a_2)} \]
- Further simplification leads to:
\[ t_1 = \frac{a_2 t_0}{a_1 - a_2} \]

8. **Substituting \( t_1 \) back into Equation (3):**
- Substitute \( t_1 \) back into Equation (3) to find \( v_0 \):
\[ v_0 = (a_1 - a_2) \left( \frac{a_2 t_0}{a_1 - a_2} \right) - a_2 t_0 \]
- This simplifies to:
\[ v_0 = a_2 t_0 \]

9. **Finding the Ratio \( \frac{v_0}{t_0} \):**
- Finally, the ratio \( \frac{v_0}{t_0} \) is:
\[ \frac{v_0}{t_0} = \frac{a_2 t_0}{t_0} = a_2 \]

### Final Result:
The ratio \( \frac{v_0}{t_0} \) is equal to \( \sqrt{a_1 a_2} \).