Multiple choice

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. The minimum required acceleration of car B to avoid an accident is

  1. 1 m $s^{-2}$
  2. 1.5 m $s^{-2}$
  3. 2 m $s^{-2}$
  4. 3 m $s^{-2}$
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A Correct answer
Explanation

Relative speed of B and A is 54 - 36 = 18 km/h = 5 m/s. Relative speed of C and A is 54 + 36 = 90 km/h = 25 m/s. Time for C to reach A is 1000m / 25 m/s = 40s. B must cover 1000m in less than 40s. Using s = ut + 0.5at^2, 1000 = 5 * 40 + 0.5 * a * 40^2. 1000 = 200 + 800a, so 800 = 800a, a = 1 m/s^2.

AI explanation

Car C approaches car A with a relative speed of 54 km/h plus 36 km/h, equaling 90 km/h, which converts to 25 m/s. The time taken for C to cover the 1 km distance to A is 1000 m divided by 25 m/s, equaling 40 seconds. For car B to overtake A in this same 40 seconds, its displacement minus A's displacement must equal 1 km; using the equation of motion for constant acceleration, (54 multiplied by 1000/3600 multiplied by 40 + 0.5 multiplied by a multiplied by 40 squared) minus (10 multiplied by 40) equals 1000. This simplifies to 600 + 800a minus 400 = 1000, leading to 800a = 800. Therefore, the minimum required acceleration of car B is 1 m/s squared.

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