On a tow lane road, car A is travelling with a speed of 36 km/h. Two cars B and C approach car A in opposite directions with a speed of 54 km/h each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. In this case, the acceleration of car B is required to avoid an accident

Speed of car $A\left(u_A\right)=36 km/h$

$=36\times \frac{5}{18}m/s \left( 1 /h=\frac{5}{18}m/s\right)$

= 10 m/s

Speed of car B and car C

$u_B=u_C=54 km/h$

$=54\times \frac{5}{18}m/s=15 m/s$

Relative velocity of car B w.r.t. car A

$u_{BA}=u_B-u_A-15-10=5 m/s$

Relative velocity of car C w.r.t. car A

$u_{CA}=u_C-u_A-15-\left(-10\right)=25 m/s$

Distance between car A and car B =1 km = 1000 m

Time taken by car C to travel distance AC = 1000 m

Time (t)= Distance/Relative velocity of car C w.r.t. car A

$=\frac{1000}{25}s=40 s$

Let car B start to accelerate with an acceleration a

Using equation of motion, 

$s=ut+\frac{1}{2}a{t}^2$

or $s=u_{BA}t+\frac{1}{2}a{t}^2$

$1000=5\times 40+\frac{1}{2}a\times {\left(40\right)}^2$

$=200+800 a$

or $800a=800$

or $a= 1 m/s^2$

Therefore, car B should accelerate with an acceleration $1 m/s^2$