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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Now, let's tackle your physics question. We have three cars involved: A, B, and C. Car A is traveling at 36 km/h, while cars B and C are approaching A from opposite directions at 54 km/h each. At a certain point, when the distances AB and AC are both 1 km, B decides to overtake A before C does.

To avoid an accident, we need to ensure that B overtakes A safely without colliding with C. This means that B needs to complete the overtaking maneuver before C reaches the same point.

To calculate the minimum acceleration required for B to overtake A safely, we can use the equations of motion.

Let's denote:

• tt as the time taken for B to overtake A,
• aBaB as the acceleration of car B.

The distance covered by B during the overtaking maneuver is the sum of the initial distance between B and A (1 km) and the distance traveled by B while overtaking A.

The distance traveled by B during the overtaking maneuver can be calculated using the equation of motion: sB=uBt+12aBt2sB=uBt+21aBt2 where:

• sBsB is the distance traveled by B,
• uBuB is the initial velocity of B (54 km/h),
• aBaB is the acceleration of B, and
• tt is the time taken.

Given that the distance traveled by B must be equal to the initial distance between A and B (1 km), we can set up the equation as follows: 1+54t+12aBt2=11+54t+21aBt2=1

We need to solve this equation for aBaB, the acceleration of car B, to find the minimum acceleration required to avoid an accident.