For both car initial speed ($$\mu $$) = 0

Let the acceleration of car A and car B is $$a$$₁ and $$a$$₂ respectively.

Also let the time taken to reach the finishing point for car A is t₁ and for car B is t₂.

Let at finishing point speed of car A is $$v$$₁ and speed of car B is $$v$$₂

According to the question,

t₂ $$-$$ t₁ = t

and   $$v$$₁ $$-$$ $$v$$₂ = $$v$$

$$ \Rightarrow $$  $$a$$₁t₁ $$-$$ $$a$$₂t₂ = $$v$$

$$ \Rightarrow $$  $$a$$₁t₁ $$-$$ $$a$$₂(t + t₁) = $$v$$ . . . . . . .(1)

As, Total distance covered by both car is equal.

So,   x_A = x_B

$$ \Rightarrow $$  $${1 \over 2}{a_1}t_1^2 = {1 \over 2}{a_2}t_2^2$$

$$ \Rightarrow $$  $$a$$₁t$$_1^2$$ = $$a$$₂ (t + t₁)²

$$ \Rightarrow $$  $$\sqrt {{a_1}} .{t_1}$$ = $$\sqrt {{a_2}} $$ . (t + t₁)

$$ \Rightarrow $$  $$\sqrt {{a_1}} .{t_1} - \sqrt {{a_2}} .{t_1} = \sqrt {{a_2}} .t$$

$$ \Rightarrow $$  t₁ = $${{\sqrt {{a_2}} .t} \over {\sqrt {{a_1}} - \sqrt {{a_2}} }}\,\,\,\,\,\,\,.....(2)$$

Now put the value of t₁ in equation (2),

($$a$$₁ $$-$$ $$a$$₂) t₁ $$-$$ $$a$$₂t = $$v$$

$$ \Rightarrow $$  (a₁ $$-$$ a₂) . $${{\sqrt {{a_2}} .t} \over {\sqrt {{a_1}} - \sqrt {{a_2}} }} - {a_2}t = v$$

$$ \Rightarrow $$  $$\left( {\sqrt {{a_1}} + \sqrt {{a_2}} } \right)\sqrt {{a_2}} .t - {a_2}t = v$$

$$ \Rightarrow $$  $$\sqrt {{a_1}{a_2}} .t + {a_2}.t - {a_2}t = v$$

$$ \Rightarrow $$  $$v$$ = $$\sqrt {{a_1}{a_2}} .t$$