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.

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: $$V_A = V_B + v \implies v = V_A - V_B$$

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:

$$S = \frac{1}{2} a_2 T_B^2 \implies T_B = \sqrt{\frac{2S}{a_2}}$$ $$V_B = a_2 T_B = a_2 \sqrt{\frac{2S}{a_2}} = \sqrt{2S a_2}$$

$$T_B = T_A + t$$ $$\sqrt{\frac{2S}{a_2}} = \sqrt{\frac{2S}{a_1}} + t$$

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 $\sqrt{2S}$:

$$\sqrt{2S} = \frac{t \sqrt{a_1 a_2}}{\sqrt{a_1} - \sqrt{a_2}} \quad (1)$$

$$v = V_A - V_B$$ $$v = \sqrt{2S a_1} - \sqrt{2S a_2}$$

Factor out $\sqrt{2S}$:

$$v = \sqrt{2S} (\sqrt{a_1} - \sqrt{a_2}) \quad (2)$$

$$v = \left( \frac{t \sqrt{a_1 a_2}}{\sqrt{a_1} - \sqrt{a_2}} \right) (\sqrt{a_1} - \sqrt{a_2})$$

The term $(\sqrt{a_1} - \sqrt{a_2})$ cancels