Let the race length be $s$.
Since both cars start from rest and move with constant accelerations:

$s = \frac{1}{2} a_1 T_1^2 = \frac{1}{2} a_2 T_2^2$

where $T_1$ and $T_2$ are the times taken by cars A and B respectively.

Use the given time difference.
Car A takes time $t$ less than car B, so

$T_2 - T_1 = t$

Relate $T_1$ and $T_2$ using equal distances.
From $a_1 T_1^2 = a_2 T_2^2$ taking positive square root,

$\sqrt{a_1} T_1 = \sqrt{a_2} T_2$

hence

$T_1 = \sqrt{\frac{a_2}{a_1}} \, T_2$

Use the time difference equation.

$T_2 - T_1 = t$ $T_2 - \sqrt{\frac{a_2}{a_1}}T_2 = t$ $T_2\left(1-\sqrt{\frac{a_2}{a_1}}\right)=t$

We do not actually need explicit $T_1, T_2$; instead, proceed to speeds.

Write finishing speeds.
Since $u=0$, final speed after time $T$ under acceleration $a$ is

$v = aT$

Therefore, $v_A = a_1 T_1, \qquad v_B = a_2 T_2$

Given that car A passes the finish with speed $\upsilon$ more than car B:

$v_A - v_B = \upsilon$

Use distance relation to simplify.
From $\sqrt{a_1}T_1 = \sqrt{a_2}T_2$ let $\sqrt{a_1}T_1 = \sqrt{a_2}T_2 = k$

Then $T_1 = \frac{k}{\sqrt{a_1}}, \qquad T_2 = \frac{k}{\sqrt{a_2}}$

Using $T_2-T_1=t$, $\frac{k}{\sqrt{a_2}} - \frac{k}{\sqrt{a_1}} = t$ $k\left(\frac{1}{\sqrt{a_2}} - \frac{1}{\sqrt{a_1}}\right)=t$

Now compute $\upsilon$.

$\upsilon = v_A-v_B = a_1T_1-a_2T_2$ $\upsilon = a_1\cdot \frac{k}{\sqrt{a_1}} - a_2\cdot \frac{k}{\sqrt{a_2}}$ $\upsilon = k(\sqrt{a_1}-\sqrt{a_2})$

From step 6, $k = \frac{t}{\frac{1}{\sqrt{a_2}}-\frac{1}{\sqrt{a_1}}}$

So, $\upsilon = \frac{t(\sqrt{a_1}-\sqrt{a_2})}{\frac{1}{\sqrt{a_2}}-\frac{1}{\sqrt{a_1}}}$

Simplify denominator: $\frac{1}{\sqrt{a_2}}-\frac{1}{\sqrt{a_1}} = \frac{\sqrt{a_1}-\sqrt{a_2}}{\sqrt{a_1a_2}}$

Therefore, $\upsilon = t(\sqrt{a_1}-\sqrt{a_2})\cdot \frac{\sqrt{a_1a_2}}{\sqrt{a_1}-\sqrt{a_2}}$ $\upsilon = t\sqrt{a_1a_2}$

Match with options.

$\boxed{\upsilon = \sqrt{a_1a_2}\, t}$

This corresponds to Option C.