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?

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Solution

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 =

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:

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