Two cars A and B start from rest and race on a straight road. Car A reaches the destination t seconds earlier than car B. If car A's speed is v greater than car B's speed, and both cars move with constant accelerations a1 and a2 respectively, find the value of v.

Step 1: Define variables and write down the given relations. Let $S$ be the total distance of the race. Both cars start from rest, so their initial velocities are $u_A = u_B = 0$. Let $a_1$ be the acceleration of car A and $a_2$ be the acceleration of car B. Let $T_A$ and $T_B$ be the times taken by car A and car B, respectively, to complete the race. Let $V_A$ and $V_B$ be the final velocities of car A and car B, respectively, at the end point.

From the problem statement:

1. Car A takes $t$ time less than car B: $T_A = T_B - t \implies T_B = T_A + t$
2. Car A's final speed is $v$ more than car B's: $V_A = V_B + v \implies v = V_A - V_B$

Step 2: Use the equations of motion for constant acceleration. For a body starting from rest ($u=0$) with constant acceleration $a$ over distance $S$ and time $T$: Distance: $S = \frac{1}{2} a T^2$ Final velocity: $V = a T$

For car A: $S = \frac{1}{2} a_1 T_A^2 \implies T_A = \sqrt{\frac{2S}{a_1}}$ $V_A = a_1 T_A = a_1 \sqrt{\frac{2S}{a_1}} = \sqrt{2S a_1}$

For car B: $S = \frac{1}{2} a_2 T_B^2 \implies T_B = \sqrt{\frac{2S}{a_2}}$ $V_B = a_2 T_B = a_2 \sqrt{\frac{2S}{a_2}} = \sqrt{2S a_2}$

Step 3: Substitute $T_A$ and $T_B$ into the time difference relation. $T_B = T_A + t$ $\sqrt{\frac{2S}{a_2}} = \sqrt{\frac{2S}{a_1}} + t$ Factor out $\sqrt{2S}$: $\sqrt{2S} \left( \frac{1}{\sqrt{a_2}} - \frac{1}{\sqrt{a_1}} \right) = t$ Combine the terms in the parenthesis: $\sqrt{2S} \left( \frac{\sqrt{a_1} - \sqrt{a_2}}{\sqrt{a_1 a_2}} \right) = t$ Solve for $\sqrt{2S}$: $\sqrt{2S} = \frac{t \sqrt{a_1 a_2}}{\sqrt{a_1} - \sqrt{a_2}} \quad (1)$

Step 4: Substitute $V_A$ and $V_B$ into the velocity difference relation. $v = V_A - V_B$ $v = \sqrt{2S a_1} - \sqrt{2S a_2}$ Factor out $\sqrt{2S}$: $v = \sqrt{2S} (\sqrt{a_1} - \sqrt{a_2}) \quad (2)$

Step 5: Substitute the expression for $\sqrt{2S}$ from equation (1) into equation (2). $v = \left( \frac{t \sqrt{a_1 a_2}}{\sqrt{a_1} - \sqrt{a_2}} \right) (\sqrt{a_1} - \sqrt{a_2})$ The term $(\sqrt{a_1} - \sqrt{a_2})$ cancels