In a car race, car A takes a time t less than car B at the finish and passes the finishing point with speed v more than that of the car B. Assuming that both the cars start from rest and travel with constant acceleration `a_1 and a_2` respectively. Show that `v=sqrt (a_1 a_2) t.`

Explanation

To solve the problem step by step, we need to analyze the motion of both cars A and B, using the information provided in the question.

### Step 1: Define Variables
Let:
- \( t_1 \) = time taken by car A to finish the race
- \( t_2 = t_1 + t \) = time taken by car B to finish the race
- \( a_1 \) = acceleration of car A
- \( a_2 \) = acceleration of car B
- \( v_2 \) = final velocity of car B
- \( v_1 = v_2 + v \) = final velocity of car A

### Step 2: Write the Equations of Motion
Since both cars start from rest, we can use the equation for final velocity:
- For car A:
\[ v_1 = a_1 t_1 \]
- For car B:
\[ v_2 = a_2 t_2 = a_2 (t_1 + t) \]

### Step 3: Relate the Velocities
From the equations above, we can express \( v_1 \) in terms of \( v_2 \):
\[ v_1 = v_2 + v \]
Substituting the expression for \( v_2 \):
\[ a_1 t_1 = a_2 (t_1 + t) + v \]

### Step 4: Rearranging the Equation
Rearranging the equation gives us:
\[ v = a_1 t_1 - a_2 (t_1 + t) \]
Expanding this:
\[ v = a_1 t_1 - a_2 t_1 - a_2 t \]
Factoring out \( t_1 \):
\[ v = (a_1 - a_2) t_1 - a_2 t \]

### Step 5: Distance Traveled by Both Cars
The distance traveled by both cars is equal since they finish the same race:
\[ \text{Distance} = \frac{1}{2} a_1 t_1^2 = \frac{1}{2} a_2 t_2^2 \]
Substituting \( t_2 = t_1 + t \):
\[ \frac{1}{2} a_1 t_1^2 = \frac{1}{2} a_2 (t_1 + t)^2 \]
Expanding the right side:
\[ \frac{1}{2} a_1 t_1^2 = \frac{1}{2} a_2 (t_1^2 + 2t_1 t + t^2) \]
Cancelling \( \frac{1}{2} \) from both sides:
\[ a_1 t_1^2 = a_2 (t_1^2 + 2t_1 t + t^2) \]

### Step 6: Rearranging for \( t_1^2 \)
Rearranging gives:
\[ a_1 t_1^2 = a_2 t_1^2 + 2 a_2 t_1 t + a_2 t^2 \]
\[ (a_1 - a_2) t_1^2 - 2 a_2 t_1 t - a_2 t^2 = 0 \]
This is a quadratic equation in \( t_1 \).

### Step 7: Solving the Quadratic Equation
Using the quadratic formula:
\[ t_1 = \frac{-(-2 a_2 t) \pm \sqrt{(-2 a_2 t)^2 - 4(a_1 - a_2)(-a_2 t^2)}}{2(a_1 - a_2)} \]
This simplifies to:
\[ t_1 = \frac{2 a_2 t \pm \sqrt{4 a_2^2 t^2 + 4 a_2 (a_1 - a_2) t^2}}{2(a_1 - a_2)} \]
Factoring out \( 2t \):
\[ t_1 = \frac{t \cdot (a_2 \pm \sqrt{a_2^2 + a_2 (a_1 - a_2)})}{(a_1 - a_2)} \]

### Step 8: Substitute \( t_1 \) back into the Velocity Equation
Substituting \( t_1 \) back into the equation for \( v \) gives:
\[ v = (a_1 - a_2) \left(\frac{t \cdot (a_2 + \sqrt{a_2^2 + a_2 (a_1 - a_2)})}{(a_1 - a_2)}\right) - a_2 t \]
After simplification, we arrive at:
\[ v = \sqrt{a_1 a_2} t \]

### Final Result
Thus, we have shown that:
\[ v = \sqrt{a_1 a_2} t \]