# Q 3.8 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?

Velocity of car A=36 km h-1 = 10 m s-1

Velocity of car B and car C = 54 km h-1 = 15 m s-1

A and B are travelling in the same direction.

A and C are travelling in opposite directions.

The velocity of A w.r.t C is VAC= 25 m s-1.

Time in which A would reach C is t

$\\t=\frac{AC}{v_{AC}}\\ =\frac{1000}{25}\\ =40s$

The velocity of B w.r.t A is VBA= 5 m s-1

Distance between A and B is s= 1000 m

Maximum time in which B has to overtake A=40s

The required acceleration can be therefore calculated using the second equation of motion

$\\s=ut+\frac{1}{2}at^{2}\\ 1000=5\times 40+\frac{1}{2}\times a\times 40^{2}\\ 1000-200=800a\\ a=1\ m\ s^{-2}$

B has to have a minimum acceleration of 1 m s-2 to avoid an accident.

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