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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, 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, vCA = vC - (- vA) = 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.

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