On a two lane road .a car A is travelling with a speed of 36km/h.two cars B and C approach car A in opposite directions with a speed of 54km.h each.At certain instant ,when the distance AB is equal to AC ,both being 1km,B decides to overtake A before C does .what minimum acceleration for car B is required to avoid accident?

Something in this question does not make sense. If A and B are travelling in the same directions, and C in the opposite direction, how can C overtake A?


when the distance AB is equal to AC ,both being 1km,B decides to overtake A before C does

If C can overtake A, as can do B, then A, B and C should be travelling in the same direction.

Or if the question originally meant that C meets A, then it would make sense.

If the latter is the case, then:

The easiest way to solve this problem is perhaps to take the speeds as relative speeds.

The relative speed of car B for an observer in car A is 36 - 54 = 18 km/h
The relative speed of car C for an observer in car A is 54 + 36 = 90 km/h

Let's be 'in car A', and from this point, it seems that the car is stationary.

Distance AB = AC = 1 km.
As car C seems to be faster, it'll take the lesser time to cover the 1 km.
Time of C = 1/90 = 40 s.

Now, car B should take less time than that.
We'll use this equation: s = ut + \frac12 at^2

1000 = \(\frac{18\times1000}{3600}\)(40) + \frac12 (a)(40)^2

Solve for a to get the acceleration of car B (in m/s^2) to meet C at car A. The acceleration of A should hence be larger than than value.