An interesting case for a race

Let the distance between starting line and the finishing line be $d$ .

Let car A finish with time $t_{1}$ and speed $v_{1}$ .

Let car B finish with time $t_{2}$ and speed $v_{2}$

We have $t_{1} - t_{2} = t$ and $v_{2} - v_{1} = v$

But,

$t_{1} = \sqrt{\frac{2d}{a_{1}}}$

$t_{2} = \sqrt{\frac{2d}{a_{2}}}$

$v_{1} = \sqrt{2a_{1}d}$

$v_{2} = \sqrt{2a_{2}d}$

Substituting the values, we get-

$t= \sqrt{2d}(\dfrac{\sqrt{a_{2}} - \sqrt{a_{1}}}{\sqrt{a_{1}a_{2}}})$

$v = \sqrt{2d}(\sqrt{a_{2}} - \sqrt{a_{1}})$

Dividing, we get-

$v = t\sqrt{a_{1}a_{2}}$

Substitute the values to get $v = 30$

Since the two distances are the same, starting from rest, at constant acc. Hence..............for time T taken by fast, we have
(1/2) * 9 * T^2 = (1/2) * 4 * (T+5)^2 ...<...> T = 10 sec.
Vel. of fast = 9 * 10, Vel. of slow car at FINISH POINT =4 * (10 + 5).
Hence v = 90 - 60 = 30.

You could have specified that "a speed v more than car B" is WHEN THE CAR B CROSSES THE LINE.

When A passes the line, it is 50 m/s faster than B.

let the speed and time of carA is "V" and"t".For B speed is "U" and time "t+5".Bot start from rest so V=9t and U=4(t+5).Now V-U=5t-20 and V+U=13t+20.Let distance be X. Now V^2= 18X U^2=8X.So V/U=3/2.Now( V+U)/(V-U)=5. place the values . u will get t=10sec. find V-U.