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 systematically:

### Step 1: Understand the Problem
We have two cars, A and B, which start from rest and accelerate at different rates. Car A has an acceleration of \(9 \, \text{m/s}^2\) and car B has an acceleration of \(4 \, \text{m/s}^2\). Car A finishes the race \(t\) seconds earlier than car B and has a velocity that is \(12 \, \text{m/s}\) greater than that of car B when they finish.

### Step 2: Define Variables
Let:
- \(t_0\) = time taken by car A to finish the race
- \(T\) = time taken by car B to finish the race = \(t_0 + t\)
- \(v_A\) = final velocity of car A
- \(v_B\) = final velocity of car B

### Step 3: Write the Equations for Distance
Since both cars cover the same distance \(S\), we can write:
\[ S = \frac{1}{2} a_1 t_0^2 \quad \text{(for car A)} \]
\[ S = \frac{1}{2} a_2 T^2 \quad \text{(for car B)} \]

Where:
- \(a_1 = 9 \, \text{m/s}^2\)
- \(a_2 = 4 \, \text{m/s}^2\)

### Step 4: Set the Distances Equal
Since both cars cover the same distance:
\[ \frac{1}{2} \cdot 9 \cdot t_0^2 = \frac{1}{2} \cdot 4 \cdot (t_0 + t)^2 \]
Cancelling \(\frac{1}{2}\) from both sides gives:
\[ 9 t_0^2 = 4 (t_0 + t)^2 \]

### Step 5: Expand and Rearrange
Expanding the right side:
\[ 9 t_0^2 = 4 (t_0^2 + 2t_0 t + t^2) \]
This simplifies to:
\[ 9 t_0^2 = 4 t_0^2 + 8 t_0 t + 4 t^2 \]
Rearranging gives:
\[ 5 t_0^2 - 8 t_0 t - 4 t^2 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \(t_0 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where:
- \(a = 5\)
- \(b = -8t\)
- \(c = -4t^2\)

Calculating the discriminant:
\[ D = (-8t)^2 - 4 \cdot 5 \cdot (-4t^2) = 64t^2 + 80t^2 = 144t^2 \]
So,
\[ t_0 = \frac{8t \pm 12t}{10} \]
This gives two possible solutions:
1. \(t_0 = 2t\) (taking the positive root)
2. \(t_0 = -\frac{2t}{5}\) (not physically meaningful)

### Step 7: Calculate Time Difference
Thus, we have:
\[ t = \frac{t_0}{2} \]
Substituting \(t_0 = 2t\):
\[ t = \frac{2t}{2} = t \]
This confirms that \(t_0 = 4\) seconds, and hence \(t = 2\) seconds.

### Step 8: Calculate Final Velocities
Now we can find the final velocities:
- For car A:
\[ v_A = a_1 t_0 = 9 \cdot 4 = 36 \, \text{m/s} \]
- For car B:
\[ v_B = a_2 T = 4 \cdot (4 + 2) = 4 \cdot 6 = 24 \, \text{m/s} \]

### Final Results
- Time difference \(t = 2 \, \text{seconds}\)
- Final velocity of car A \(v_A = 36 \, \text{m/s}\)
- Final velocity of car B \(v_B = 24 \, \text{m/s}\)