Physics-kinematics question, Please Help, velocity-acceleration type

on a 2 lane road, car A is traveling with a speed of 10m/s.two cars B and C approach car A in opposite directions with speed of 15 m /s each.At a certain instant when the distance AB=AC both being 1ooom, B decides to overtake A before C . What is the minimum acceleration required to avoid an accident.??

B and A must be going in the same direction as B is trying to overtake A.
Start by drawing a clear diagram, soty of like this:

B>_________A>________ 15m/s_____10m/s______15m/s
<----1000m---><____1000m>

Step 1. Find how far A has travelled (x) when level with C, and how long it took (t)
A travels x. B travels 1000-x. Time taken for both = t. t = distance /speed
For A: t = x / 10
For B: t = 1000-x /15
x/10 = (1000-x)/15
15x = 10,000 -10x
25x = 10,000
x = 400m
t = x/10 = 40s

Step 2
Find A's acceleration
A must travel 1000m + 400m in less than time t, so that it has passed A before A nd C are level.
s= ut+at^2 / 2
1400 = 15x40 + a x 40^2 / 2
= 600 + 800a
a = (1400-600)/800 = 1m/s^2

So this is the minimum acceleration

To avoid an accident, the cars must come to a standstill;

v = 15+10 = 25 m/s; (Relative velocity);
s = 1000 metres; (Distance);
a = Deceleration Required;

a = v×v/(2×s) = 0.3125 m/sq.s;

So, acceleration = -0.3125 m/sq.s --- --- --- (ans);

that's 4m H.C. Verma .. :D