Two cars travelling towards each other on a straight road at velocity `10 m//s` and `12 m//s` respectively. When they are 150 metre apart, both drivers apply their brakes and each car decelerates at `2 m//s^(2)` until it stops. How far apart will they be when they have both come to a stop?

Options

A. 89 m
B. 98 m
C. 108 m
D. 150 m

Answer

To solve the problem step by step, we will use the equations of motion to determine how far each car travels before coming to a stop, and then calculate the distance between them when they have both stopped.

### Step 1: Identify the given data
- Velocity of Car A (v_A) = 10 m/s
- Velocity of Car B (v_B) = 12 m/s
- Initial distance between the cars (d_initial) = 150 m
- Deceleration of both cars (a) = -2 m/s² (negative because it's deceleration)

### Step 2: Calculate the stopping distance for Car A
Using the third equation of motion:
\[ v^2 = u^2 + 2as \]
Where:
- v = final velocity = 0 m/s (when the car stops)
- u = initial velocity = 10 m/s
- a = -2 m/s² (deceleration)
- s = stopping distance

Substituting the values:
\[ 0 = (10)^2 + 2(-2)s \]
\[ 0 = 100 - 4s \]
\[ 4s = 100 \]
\[ s_A = \frac{100}{4} = 25 \text{ m} \]

### Step 3: Calculate the stopping distance for Car B
Using the same equation for Car B:
\[ v^2 = u^2 + 2as \]
Where:
- v = final velocity = 0 m/s
- u = initial velocity = 12 m/s
- a = -2 m/s²

Substituting the values:
\[ 0 = (12)^2 + 2(-2)s \]
\[ 0 = 144 - 4s \]
\[ 4s = 144 \]
\[ s_B = \frac{144}{4} = 36 \text{ m} \]

### Step 4: Calculate the distance between the two cars when they stop
The total distance covered by both cars is:
\[ s_A + s_B = 25 \text{ m} + 36 \text{ m} = 61 \text{ m} \]

Now, we subtract this distance from the initial distance:
\[ d_{final} = d_{initial} - (s_A + s_B) \]
\[ d_{final} = 150 \text{ m} - 61 \text{ m} = 89 \text{ m} \]

### Conclusion
The distance between the two cars when they have both come to a stop is **89 meters**.

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