A and B participate in a race with acceleration `a_1 and a_2` , respectively . A reaches t times earlier than B at finish line and their velocities at finish line are `v_1 and v_2` , respectively. If difference between their velocities is v , then find the value of v

Options

A. `(a_1+a_2)/2t`
B. `sqrt(a_1a_2)t`
C. `(a_1a_2)/(a_1+a_2)t`
D. `(2a_1a_2)/(a_1+a_2)t`

Answer

To solve the problem step by step, we will analyze the motion of both A and B in the race, using the equations of motion and the information provided.

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

### Step 2: Write the Equations for Final Velocities
Using the equations of motion, the final velocities can be expressed as:
- For A:
\[ v_1 = a_1 t_1 \]
- For B:
\[ v_2 = a_2 t_2 = a_2 (t_1 + t) \]

### Step 3: Set Up the Equation for Velocity Difference
From the problem, we know that:
\[ v = v_1 - v_2 \]
Substituting the expressions for \( v_1 \) and \( v_2 \):
\[ v = a_1 t_1 - a_2 (t_1 + t) \]
This simplifies to:
\[ v = a_1 t_1 - a_2 t_1 - a_2 t \]
\[ v = (a_1 - a_2) t_1 - a_2 t \]

### Step 4: Write the Equations for Distances
Since both A and B cover the same distance \( S \):
- For A:
\[ S = \frac{1}{2} a_1 t_1^2 \]
- For B:
\[ S = \frac{1}{2} a_2 t_2^2 = \frac{1}{2} a_2 (t_1 + t)^2 \]

### Step 5: Equate the Distances
Equating the distances gives:
\[ \frac{1}{2} a_1 t_1^2 = \frac{1}{2} a_2 (t_1 + t)^2 \]
Cancelling \( \frac{1}{2} \) from both sides:
\[ a_1 t_1^2 = a_2 (t_1^2 + 2t t_1 + t^2) \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ a_1 t_1^2 = a_2 t_1^2 + 2 a_2 t t_1 + a_2 t^2 \]
\[ (a_1 - a_2) t_1^2 - 2 a_2 t t_1 - a_2 t^2 = 0 \]

### Step 7: Solve for \( t_1 \)
This is a quadratic equation in \( t_1 \):
\[ (a_1 - a_2) t_1^2 - 2 a_2 t t_1 - a_2 t^2 = 0 \]
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:
\[ t_1 = \frac{2 a_2 t \pm \sqrt{4 a_2^2 t^2 + 4 a_2 t^2 (a_1 - a_2)}}{2(a_1 - a_2)} \]
\[ t_1 = \frac{a_2 t \pm \sqrt{a_2^2 t^2 + a_2 t^2 (a_1 - a_2)}}{(a_1 - a_2)} \]

### Step 8: Substitute \( t_1 \) Back to Find \( v \)
Now substitute \( t_1 \) back into the equation for \( v \):
\[ v = (a_1 - a_2) t_1 - a_2 t \]
After simplification, we find:
\[ v = \sqrt{a_1 a_2} t \]

### Final Result
Thus, the difference in their velocities is:
\[ \boxed{v = \sqrt{a_1 a_2} t} \]