Two cars start in a race with velocities `u_(1)` and `u_(2)` and travel in a straight line with acceleration 'a' and b .If both reach the finish line at the same time,the range of the race is

Options

A. `(2(u_1 - u_2))/((beta - alpha)^2) (u_1 beta - u_2 alpha)`
B. `(2(u_1 - u_2))/(beta + alpha) (u_1 alpha - u_2 beta)`
C. `(2(u_1 - u_2)^2)/((beta - alpha)^2) `
D. `(2 u_1 u_2)/(beta alpha)`

Answer

To solve the problem, we need to find the range of the race for two cars starting with different initial velocities and accelerations, reaching the finish line at the same time. We will use the equations of motion to derive the solution step by step.

### Step-by-Step Solution:

1. **Identify the Given Variables:**
- Let the initial velocity of the first car be \( u_1 \).
- Let the initial velocity of the second car be \( u_2 \).
- Let the acceleration of the first car be \( a \).
- Let the acceleration of the second car be \( b \).
- Let the distance of the race (range) be \( x \).
- Let the time taken to finish the race be \( t \).

2. **Write the Equation of Motion for Both Cars:**
- For the first car, the equation of motion is:
\[ x = u_1 t + \frac{1}{2} a t^2 \quad \text{(Equation 1)} \]
- For the second car, the equation of motion is:
\[ x = u_2 t + \frac{1}{2} b t^2 \quad \text{(Equation 2)} \]

3. **Set the Distances Equal:**
Since both cars reach the finish line at the same time, we can set the two equations equal to each other:
\[ u_1 t + \frac{1}{2} a t^2 = u_2 t + \frac{1}{2} b t^2 \]

4. **Rearranging the Equation:**
Rearranging gives us:
\[ u_1 t - u_2 t = \frac{1}{2} b t^2 - \frac{1}{2} a t^2 \]
Simplifying further:
\[ (u_1 - u_2) t = \frac{1}{2} (b - a) t^2 \]

5. **Dividing by \( t \) (assuming \( t \neq 0 \)):**
\[ u_1 - u_2 = \frac{1}{2} (b - a) t \]
From this, we can express \( t \):
\[ t = \frac{2(u_1 - u_2)}{b - a} \quad \text{(Equation 3)} \]

6. **Substituting \( t \) back into one of the equations:**
We can substitute \( t \) from Equation 3 into Equation 1 to find \( x \):
\[ x = u_1 \left(\frac{2(u_1 - u_2)}{b - a}\right) + \frac{1}{2} a \left(\frac{2(u_1 - u_2)}{b - a}\right)^2 \]

7. **Calculating \( x \):**
- The first term becomes:
\[ x_1 = \frac{2u_1(u_1 - u_2)}{b - a} \]
- The second term becomes:
\[ x_2 = \frac{1}{2} a \cdot \frac{4(u_1 - u_2)^2}{(b - a)^2} = \frac{2a(u_1 - u_2)^2}{(b - a)^2} \]
- Therefore, the total distance \( x \) is:
\[ x = \frac{2u_1(u_1 - u_2)}{b - a} + \frac{2a(u_1 - u_2)^2}{(b - a)^2} \]

8. **Final Expression for the Range:**
\[ x = \frac{2(u_1 - u_2)}{b - a} \left( u_1 + \frac{a(u_1 - u_2)}{b - a} \right) \]