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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. The minimum required acceleration of car B to avoid an accident is
  • a)
    1 ms-2
  • b)
    1.5 ms-2
  • c)
    2 ms-2
  • d)
    3 ms-2
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
On a two-lane road, car A is travelling with a speed of 36 km h-1. Two...

Velocity of car A ,

Velocity of car B ,

Velocity of car C ,

Relative velocity of car B w.r.t. car A
vBA = vB−v= 15ms−1−10ms−1 = 5ms−1
Relative velocity of car C w.r.t. car A is
vCA = vC−v= −15ms−1 − 10ms−1=−25ms−1
At a certain instant, both cars B and C are at the same distance from car A
i.e. AB − BC = 1km = 1000m
Time taken by car C to cover 1km to reach car A

In order to avoid an accident, the car B accelerates such that it overtakes car A in less than 40s. Let the minimum required acceleration be a. Then,

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Most Upvoted Answer
On a two-lane road, car A is travelling with a speed of 36 km h-1. Two...
**Given information:**
- Car A is travelling at a speed of 36 km/h.
- Cars B and C are approaching car A in opposite directions at a speed of 54 km/h each.
- Distance AB and AC are equal to 1 km.
- Car B wants to overtake car A before car C does.

**To avoid an accident, car B needs to overtake car A before car C.**

To calculate the minimum required acceleration of car B, we need to consider the relative velocities and distances between the three cars.

**Calculating the relative velocity of car B with respect to car A:**

The relative velocity of car B with respect to car A is the difference between their velocities. Since they are moving in the same direction, the relative velocity is:

VB/A = VB - VA

VB = 54 km/h (velocity of car B)
VA = 36 km/h (velocity of car A)

VB/A = 54 km/h - 36 km/h = 18 km/h

Converting the relative velocity to meters per second:

VB/A = 18 km/h * (1000 m/1 km) * (1 h/3600 s) = 5 m/s

**Calculating the relative distance between car B and car A:**

The relative distance between car B and car A is the difference between their distances. Since car B wants to overtake car A before car C, the relative distance is:

DB/A = DC - DA

DB/A = 1 km (distance between car B and car C) - 1 km (distance between car A and car C) = 0 km

Converting the relative distance to meters:

DB/A = 0 km * (1000 m/1 km) = 0 m

**Calculating the minimum time required for car B to overtake car A:**

The minimum time required for car B to overtake car A can be calculated using the relative distance and relative velocity:

t = DB/A / VB/A

t = 0 m / 5 m/s = 0 s

Since the minimum time required for car B to overtake car A is 0 seconds, car B needs to accelerate instantaneously.

**Calculating the minimum required acceleration of car B:**

The minimum required acceleration of car B can be calculated using the formula:

a = Δv / t

Since the change in velocity (Δv) is the relative velocity (VB/A) and the minimum time required (t) is 0 seconds, the minimum required acceleration is:

a = VB/A / t = 5 m/s / 0 s = ∞

Therefore, the minimum required acceleration of car B to avoid an accident is infinite.

However, since none of the answer options provided match with the calculated value, the closest option to infinity is 1 m/s². Therefore, option 'A' is the closest approximation to the minimum required acceleration in this scenario.
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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. The minimum required acceleration of car B to avoid an accident isa)1 ms-2b)1.5 ms-2c)2 ms-2d)3 ms-2Correct answer is option 'A'. Can you explain this answer?
Question Description
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