Two cars are travelling towards each other on a straight road at velocities `15 m//s` and `m//s` respectively. When they are 150 m apart both the drivers apply the brakes and the car decelerate at `3 m//s^(2)` and `4 m//s^(2)` until they stop separation between the cars when they come to rest is

Options

A. 86.5 m
B. 89.5 m
C. 85.5 m
D. 80.5 m

Explanation

To solve the problem, we need to determine the separation between two cars after they come to a stop. Let's break down the solution step by step.

### Step 1: Identify the initial conditions
- Car A has an initial velocity \( u_1 = 15 \, \text{m/s} \) and decelerates at \( a_1 = -3 \, \text{m/s}^2 \).
- Car B has an initial velocity \( u_2 = 16 \, \text{m/s} \) and decelerates at \( a_2 = -4 \, \text{m/s}^2 \).
- The initial distance between the two cars is \( d = 150 \, \text{m} \).

### Step 2: Calculate the stopping distance for Car A
Using the formula:
\[ v^2 = u^2 + 2as \]
where \( v \) is the final velocity (0 m/s when the car stops), \( u \) is the initial velocity, \( a \) is the acceleration (deceleration in this case), and \( s \) is the stopping distance.

For Car A:
\[ 0 = (15)^2 + 2(-3)s_1 \]
\[ 0 = 225 - 6s_1 \]
\[ 6s_1 = 225 \]
\[ s_1 = \frac{225}{6} = 37.5 \, \text{m} \]

### Step 3: Calculate the stopping distance for Car B
Using the same formula for Car B:
\[ 0 = (16)^2 + 2(-4)s_2 \]
\[ 0 = 256 - 8s_2 \]
\[ 8s_2 = 256 \]
\[ s_2 = \frac{256}{8} = 32 \, \text{m} \]

### Step 4: Calculate the total distance covered by both cars
The total distance covered by both cars when they stop is:
\[ s_{total} = s_1 + s_2 = 37.5 \, \text{m} + 32 \, \text{m} = 69.5 \, \text{m} \]

### Step 5: Calculate the final separation between the cars
The final separation \( S \) between the two cars when they come to rest is given by:
\[ S = d - s_{total} = 150 \, \text{m} - 69.5 \, \text{m} = 80.5 \, \text{m} \]

### Final Answer
The separation between the cars when they come to rest is **80.5 m**.

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