### Q.8:- 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. What minimum acceleration of car B is required to avoid an accident ?

**Answer:-**

Velocity of car A, *v*_{A} = 36 km/h = 10 m/s

Velocity of car B, *v*_{B} = 54 km/h = 15 m/s

Velocity of car C, *v*_{C} = 54 km/h = 15 m/s

Relative velocity of car B with respect to car A,

*v*_{BA} = *v*_{B} – *v*_{A}

= 15 – 10 = 5 m/s

Relative velocity of car C with respect to car A,

*v*_{CA} = *v*_{C }– (– *v*_{A})

= 15 + 10 = 25 m/s

At a certain instance, both cars B and C are at the same distance from car A i.e.,

*s* = 1 km = 1000 m

Time taken (*t*) by car C to cover 1000 m = 1000 / 25 = 40 s

Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s.

From second equation of motion, minimum acceleration (*a*) produced by car B can be obtained as:

s = ut + (1/2)at^{2}

1000 = 5 × 40 + (1/2) × a × (40)^{2}

a = 1600 / 1600 = 1 ms^{-2}