A driver having a definite reaction time is capable of stopping his car over a distance of 30 m on seeing a red traffic signal, when the speed of the car is 72 km/hr andover a distance of 10 m when the speed is 36 km/hr. Find the distance over which he can stop the car if it were running at a speed of 54 km/hr. Assume that his reaction time and the deceleration of the car remains same in all the three cases.

Explanation

To solve the problem, we need to analyze the stopping distances of the car at different speeds and relate them to the driver's reaction time and the car's deceleration.

### Given Data:
1. **Case 1**:
- Speed = 72 km/hr = 20 m/s
- Stopping distance (total) = 30 m

2. **Case 2**:
- Speed = 36 km/hr = 10 m/s
- Stopping distance (total) = 10 m

3. **Case 3**:
- Speed = 54 km/hr = 15 m/s
- Stopping distance (total) = D (unknown)

### Step 1: Define Variables
Let:
- \( t \) = reaction time (in seconds)
- \( a \) = deceleration (in m/s²)

### Step 2: Analyze Each Case
For each case, the total stopping distance \( D \) can be expressed as:
\[ D = x_1 + x_2 \]
where:
- \( x_1 = v \cdot t \) (distance traveled during reaction time)
- \( x_2 = \frac{v^2}{2a} \) (distance traveled while decelerating to rest)

### Case 1:
1. **For 72 km/hr (20 m/s)**:
- \( x_1 = 20t \)
- \( x_2 = \frac{(20)^2}{2a} = \frac{400}{2a} = \frac{200}{a} \)
- Total distance:
\[ 20t + \frac{200}{a} = 30 \quad \text{(Equation 1)} \]

### Case 2:
2. **For 36 km/hr (10 m/s)**:
- \( x_1 = 10t \)
- \( x_2 = \frac{(10)^2}{2a} = \frac{100}{2a} = \frac{50}{a} \)
- Total distance:
\[ 10t + \frac{50}{a} = 10 \quad \text{(Equation 2)} \]

### Step 3: Solve Equations
Now we have two equations:
1. \( 20t + \frac{200}{a} = 30 \)
2. \( 10t + \frac{50}{a} = 10 \)

From Equation 2, we can express \( \frac{50}{a} \):
\[ \frac{50}{a} = 10 - 10t \quad \Rightarrow \quad \frac{1}{a} = \frac{10 - 10t}{50} = \frac{1 - t}{5} \]

Substituting \( \frac{1}{a} \) into Equation 1:
\[ 20t + 200 \cdot \frac{1 - t}{5} = 30 \]
Simplifying:
\[ 20t + 40(1 - t) = 30 \]
\[ 20t + 40 - 40t = 30 \]
\[ -20t + 40 = 30 \]
\[ -20t = -10 \quad \Rightarrow \quad t = 0.5 \text{ seconds} \]

Now substituting \( t \) back to find \( a \):
\[ \frac{50}{a} = 10 - 10(0.5) = 5 \quad \Rightarrow \quad a = \frac{50}{5} = 10 \text{ m/s}^2 \]

### Step 4: Analyze Case 3
Now we can find the stopping distance for the speed of 54 km/hr (15 m/s):
- \( x_1 = 15t = 15 \times 0.5 = 7.5 \)
- \( x_2 = \frac{(15)^2}{2a} = \frac{225}{20} = 11.25 \)

Total stopping distance \( D \):
\[ D = x_1 + x_2 = 7.5 + 11.25 = 18.75 \text{ m} \]

### Final Answer:
The distance over which the driver can stop the car at a speed of 54 km/hr is **18.75 meters**.