On a two-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. What minimum acceleration of car B is required to avoid an accident?
Given,
Velocity of car A, vA = 36 km/h = 10 m/s
Velocity of car B, vB = 54 km/h = 15 m/s
Velocity of car C, vC = 54 km/h = 15 m/s
Relative velocity of car B with respect to car A,
vBA = vB – vA = 15-10 = 5 m/s
Relative velocity of car C with respect to car A,
vCB = vC – (-vA)= 15+10 = 25 m/s
At a certain time, both cars B and C are at the same distance,
s = 1 km = 1000m
Time taken (t) by car C to just reach car A is,
t = 
= 40 s
∴ The minimum acceleration (a) required for car B to just beat car C in reaching car A is given by,
From 2nd equation of motion,
s = ut + 0.5at2
where,
u = Initial velocity = 5 m/s
a = Acceleration/Deceleration
s = Distance covered = 1000 m
t = Time = 40 s
⇒ 1000 = (5×40) + (0.5×a×402)
∴ a = 
ms-2 = 1 ms-2
Hence, car B require minimum acceleration, a = 1 ms-2 to beat car C.