In a car race, car `A` takes `4 s` less than can `B` at the finish and passes the finishing point with a velcity `v` more than the car `B` . Assumung that the cars start form restand travel with constant accleration `a_(1)=4 m s^(-2)` and `a_(2) =1 m s^(-2)` respectively, find the velocity of `v` in m `s^(-1)`.

Explanation

To solve the problem, we need to find the velocity \( v \) of car A in meters per second. We know the following:

1. Car A has an acceleration \( a_1 = 4 \, \text{m/s}^2 \).
2. Car B has an acceleration \( a_2 = 1 \, \text{m/s}^2 \).
3. Car A finishes 4 seconds earlier than car B.
4. Both cars start from rest.

### Step-by-Step Solution:

**Step 1: Define the time taken by car A and car B.**

Let the time taken by car A to finish the race be \( t_0 \). Then, the time taken by car B will be \( t_0 + 4 \).

**Step 2: Write the equations for distance traveled by both cars.**

Since both cars travel the same distance \( s \), we can use the formula for distance under constant acceleration:

\[ s = \frac{1}{2} a t^2 \]

For car A:
\[ s = \frac{1}{2} a_1 t_0^2 = \frac{1}{2} \cdot 4 \cdot t_0^2 = 2 t_0^2 \]

For car B:
\[ s = \frac{1}{2} a_2 (t_0 + 4)^2 = \frac{1}{2} \cdot 1 \cdot (t_0 + 4)^2 = \frac{1}{2} (t_0 + 4)^2 \]

**Step 3: Set the distances equal to each other.**

Since both distances are equal, we have:

\[ 2 t_0^2 = \frac{1}{2} (t_0 + 4)^2 \]

**Step 4: Solve for \( t_0 \).**

Multiply both sides by 2 to eliminate the fraction:

\[ 4 t_0^2 = (t_0 + 4)^2 \]

Expanding the right side:

\[ 4 t_0^2 = t_0^2 + 8 t_0 + 16 \]

Rearranging gives:

\[ 4 t_0^2 - t_0^2 - 8 t_0 - 16 = 0 \]

This simplifies to:

\[ 3 t_0^2 - 8 t_0 - 16 = 0 \]

**Step 5: Use the quadratic formula to solve for \( t_0 \).**

The quadratic formula is given by:

\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 3 \), \( b = -8 \), and \( c = -16 \):

\[ t_0 = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-16)}}{2 \cdot 3} \]

Calculating the discriminant:

\[ t_0 = \frac{8 \pm \sqrt{64 + 192}}{6} = \frac{8 \pm \sqrt{256}}{6} = \frac{8 \pm 16}{6} \]

This gives two potential solutions:

1. \( t_0 = \frac{24}{6} = 4 \)
2. \( t_0 = \frac{-8}{6} \) (not valid since time cannot be negative)

Thus, \( t_0 = 4 \, \text{s} \).

**Step 6: Calculate the velocities of both cars.**

For car A:
\[ v_1 = a_1 \cdot t_0 = 4 \cdot 4 = 16 \, \text{m/s} \]

For car B:
\[ v_2 = a_2 \cdot (t_0 + 4) = 1 \cdot (4 + 4) = 8 \, \text{m/s} \]

**Step 7: Find the velocity \( v \).**

According to the problem, car A passes the finishing point with a velocity \( v \) more than car B:

\[ v = v_1 - v_2 = 16 - 8 = 8 \, \text{m/s} \]

### Final Answer:
The velocity \( v \) is \( 8 \, \text{m/s} \).