In a car race, car A takes time t less than car B and passes the finishing point with a velocity of `12m//s` more than the velocity with which car B passes the finishing point. Assume that the cars A and B start from rest and travel with constant acceleration of `9 m//s^(2) `and ` 4 m//s^(2)` , respectively. If `v_(A)` and `v_(B)` be the velocities of cars A and B, respectively, then

Options

A. t= 2 sec
B. t = 3 sec
C. `v_(A) = 24m//s, v_(B) = 12m//s`
D. `v_(A) = 36m//s, v_(B) = 24m//s`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the Variables
Let:
- \( v_A \) = final velocity of car A
- \( v_B \) = final velocity of car B
- \( t \) = time taken by car A less than car B
- \( T \) = time taken by car A to finish the race
- \( T + t \) = time taken by car B to finish the race

### Step 2: Use the Equations of Motion
For car A:
- Initial velocity \( u_A = 0 \)
- Acceleration \( a_A = 9 \, m/s^2 \)
- Final velocity \( v_A = u_A + a_A T = 0 + 9T = 9T \)

For car B:
- Initial velocity \( u_B = 0 \)
- Acceleration \( a_B = 4 \, m/s^2 \)
- Final velocity \( v_B = u_B + a_B (T + t) = 0 + 4(T + t) = 4T + 4t \)

### Step 3: Relate the Velocities
According to the problem, car A passes the finishing point with a velocity of \( 12 \, m/s \) more than car B:
\[ v_A = v_B + 12 \]
Substituting the expressions for \( v_A \) and \( v_B \):
\[ 9T = (4T + 4t) + 12 \]
Simplifying this equation:
\[ 9T = 4T + 4t + 12 \]
\[ 5T - 4t = 12 \quad \text{(Equation 1)} \]

### Step 4: Use the Distance Formula
The distance traveled by both cars is the same. Using the equation of motion for distance:
For car A:
\[ S = u_A T + \frac{1}{2} a_A T^2 = 0 + \frac{1}{2} \cdot 9 \cdot T^2 = \frac{9}{2} T^2 \]

For car B:
\[ S = u_B (T + t) + \frac{1}{2} a_B (T + t)^2 = 0 + \frac{1}{2} \cdot 4 \cdot (T + t)^2 = 2(T + t)^2 \]

Setting the distances equal:
\[ \frac{9}{2} T^2 = 2(T + t)^2 \]
Expanding and simplifying:
\[ 9T^2 = 4(T^2 + 2Tt + t^2) \]
\[ 9T^2 = 4T^2 + 8Tt + 4t^2 \]
\[ 5T^2 - 8Tt - 4t^2 = 0 \quad \text{(Equation 2)} \]

### Step 5: Solve the System of Equations
Now we have two equations:
1. \( 5T - 4t = 12 \)
2. \( 5T^2 - 8Tt - 4t^2 = 0 \)

From Equation 1, we can express \( T \) in terms of \( t \):
\[ T = \frac{12 + 4t}{5} \]

Substituting this into Equation 2:
\[ 5\left(\frac{12 + 4t}{5}\right)^2 - 8\left(\frac{12 + 4t}{5}\right)t - 4t^2 = 0 \]
Simplifying this will give us a quadratic equation in terms of \( t \).

### Step 6: Calculate Velocities
Once we find \( t \), we can substitute back to find \( T \) and then calculate \( v_A \) and \( v_B \):
\[ v_A = 9T \]
\[ v_B = 4(T + t) \]

### Final Values
After solving the equations, we find:
- \( v_A = 36 \, m/s \)
- \( v_B = 24 \, m/s \)