On a two lane road , car (A) is travelling with a speed of `36 km h^(-1)`. Tho car ` 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 accelration of car (B) is required to avoid and accident.

Options

A. `9.8 m s^(-2)`
B. `10 m s^(-2)`
C. `1 m s^(-2)`
D. `2.0 m s^(-2)`

Answer

To solve the problem step by step, we will analyze the motion of the cars involved and use the equations of motion to determine the minimum acceleration required for car B to overtake car A before car C reaches car A.

### Step 1: Convert speeds from km/h to m/s
- Speed of car A = 36 km/h = \( \frac{36 \times 1000}{3600} = 10 \) m/s
- Speed of car B = 54 km/h = \( \frac{54 \times 1000}{3600} = 15 \) m/s
- Speed of car C = 54 km/h = \( \frac{54 \times 1000}{3600} = 15 \) m/s

### Step 2: Determine the relative speed of car C with respect to car A
- The relative speed of car C with respect to car A = Speed of C + Speed of A = \( 15 + 10 = 25 \) m/s

### Step 3: Calculate the time taken by car C to reach car A
- Distance between car A and car C = 1000 m
- Time taken by car C to reach car A = \( \frac{\text{Distance}}{\text{Relative Speed}} = \frac{1000}{25} = 40 \) seconds

### Step 4: Determine the relative speed of car B with respect to car A
- The relative speed of car B with respect to car A = Speed of B - Speed of A = \( 15 - 10 = 5 \) m/s

### Step 5: Calculate the time taken by car B to reach car A
Let \( t \) be the time taken by car B to reach car A. The distance between car A and car B is also 1000 m. Using the formula:
\[ \text{Distance} = \text{Relative Speed} \times t \]
We have:
\[ 1000 = 5t \]
Thus,
\[ t = \frac{1000}{5} = 200 \text{ seconds} \]

### Step 6: Since car C takes 40 seconds to reach car A, car B must reach car A in less than 40 seconds
- Therefore, car B must accelerate to cover the distance of 1000 m in 40 seconds.

### Step 7: Use the equation of motion to find the required acceleration
Using the equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
Where:
- \( S = 1000 \) m (distance to cover)
- \( u = 5 \) m/s (initial speed of car B relative to car A)
- \( t = 40 \) s (time to reach car A)

Substituting the values:
\[ 1000 = 5 \times 40 + \frac{1}{2} a \times (40^2) \]
\[ 1000 = 200 + \frac{1}{2} a \times 1600 \]
\[ 1000 - 200 = 800 = \frac{1}{2} a \times 1600 \]
\[ 800 = 800a \]
\[ a = 1 \text{ m/s}^2 \]

### Conclusion
The minimum acceleration required for car B to avoid an accident and overtake car A before car C reaches car A is \( 1 \text{ m/s}^2 \).