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. `1 m s^(-2)`
B. `1.5 m s^(-2)`
C. `2 m s^(-2)`
D. `3 m s^(-2)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert speeds from km/h to m/s
- The speed of car A is given as 36 km/h.
- To convert this to m/s, we use the conversion factor \( \frac{5}{18} \):
\[ v_A = 36 \times \frac{5}{18} = 10 \, \text{m/s} \]

- The speed of cars B and C is given as 54 km/h.
- Converting this to m/s:
\[ v_B = v_C = 54 \times \frac{5}{18} = 15 \, \text{m/s} \]

### Step 2: Calculate the relative speed of car C with respect to car A
- Since car C is approaching from the opposite direction, the relative speed of car C with respect to car A is:
\[ v_{rel} = v_A + v_C = 10 + 15 = 25 \, \text{m/s} \]

### Step 3: Calculate the time taken by car C to reach car A
- The distance between car A and car C is 1 km (1000 m). The time taken by car C to reach car A can be calculated using:
\[ t = \frac{d}{v_{rel}} = \frac{1000}{25} = 40 \, \text{seconds} \]

### Step 4: Calculate the distance traveled by car A in that time
- The distance traveled by car A in 40 seconds is:
\[ d_A = v_A \times t = 10 \times 40 = 400 \, \text{m} \]

### Step 5: Calculate the total distance car B needs to cover to overtake car A
- The total distance car B needs to cover to overtake car A before car C reaches is:
\[ d_{total} = 1000 \, \text{m} + 400 \, \text{m} = 1400 \, \text{m} \]

### Step 6: Use the kinematic equation to find the minimum acceleration of car B
- We will use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s = 1400 \, \text{m} \)
- \( u = 15 \, \text{m/s} \)
- \( t = 40 \, \text{s} \)

- Plugging in the values:
\[ 1400 = 15 \times 40 + \frac{1}{2} a (40^2) \]
\[ 1400 = 600 + 800a \]
\[ 800a = 1400 - 600 \]
\[ 800a = 800 \]
\[ a = \frac{800}{800} = 1 \, \text{m/s}^2 \]

### Final Answer
- The minimum acceleration of car B required to avoid an accident is \( 1 \, \text{m/s}^2 \).

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