On a two-lane road, car A is travelling with a speed of 36 km h^-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h^-1 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. The minimum acceleration(m/s²) of  car B is required to avoid an accident is $\_\_\_\_\_\_$

Speed of A = 36km/hr = 36 × 5/18​m/s = 10m/s

Speed of B = Speed of C = 54 × 5/18​m/s = 15m/s

Relative speed of A w.r.t C = 10+15 = 25m/s

Time taken by C to overtake A = 1000​/25

= 40sec.

Distance traveled by A all this time = 10×40 = 400m

So, B has to cover the distance of (1000m+400m)

i.e. 1400m to take over a A before C does. in 40sec.

Now putting in formula ;-

s = ut + 1/2​at²

1400 = 15×40 + 1/2​×a×(40)²

or, 800 = a × 800

a=1m/s²