The driver of a car moving with a speed of `10 ms^(-1) ` sees a red light ahead, applies brakes and stops after covering 10 m distance. If the same car were moving with a speed of `20 ms^(-1)` the same driver would have stopped the car after covering 30 m distance. Within what distance the car can be stopped if traveling with a velocity of `15 ms^(-1) ` Assume the same reaction time and the same deceleration in each case

Options

A. ` 18.75 m`
B. ` 20.75`
C. ` 22.75m`
D. ` 25 m`

Explanation

To solve the problem, we need to analyze the stopping distances of the car at different speeds and then find the stopping distance at a speed of 15 m/s.

### Step 1: Identify the known variables
- Speed when the car stops after 10 m: \( U_1 = 10 \, \text{m/s} \)
- Stopping distance at \( U_1 \): \( S_1 = 10 \, \text{m} \)
- Speed when the car stops after 30 m: \( U_2 = 20 \, \text{m/s} \)
- Stopping distance at \( U_2 \): \( S_2 = 30 \, \text{m} \)
- Speed for which we need to find the stopping distance: \( U_3 = 15 \, \text{m/s} \)

### Step 2: Use the equations of motion
The stopping distance can be expressed in terms of initial speed, reaction time, and deceleration. The total stopping distance \( S \) can be given by:
\[ S = U \cdot T_r + \frac{U^2}{2a} \]
Where:
- \( U \) is the initial speed,
- \( T_r \) is the reaction time,
- \( a \) is the deceleration.

### Step 3: Set up equations for the first two cases
For the first case (10 m/s):
\[ S_1 = U_1 \cdot T_r + \frac{U_1^2}{2a} \]
\[ 10 = 10 \cdot T_r + \frac{10^2}{2a} \]
\[ 10 = 10T_r + \frac{100}{2a} \]
\[ 10 = 10T_r + \frac{50}{a} \quad \text{(Equation 1)} \]

For the second case (20 m/s):
\[ S_2 = U_2 \cdot T_r + \frac{U_2^2}{2a} \]
\[ 30 = 20 \cdot T_r + \frac{20^2}{2a} \]
\[ 30 = 20T_r + \frac{400}{2a} \]
\[ 30 = 20T_r + \frac{200}{a} \quad \text{(Equation 2)} \]

### Step 4: Solve the equations simultaneously
From Equation 1:
\[ 10 - 10T_r = \frac{50}{a} \]
\[ a = \frac{50}{10 - 10T_r} \quad \text{(Equation 3)} \]

Substituting Equation 3 into Equation 2:
\[ 30 = 20T_r + \frac{200}{\frac{50}{10 - 10T_r}} \]
\[ 30 = 20T_r + \frac{200(10 - 10T_r)}{50} \]
\[ 30 = 20T_r + 40 - 40T_r \]
\[ 30 = 40 - 20T_r \]
\[ 20T_r = 10 \]
\[ T_r = 0.5 \, \text{s} \]

### Step 5: Substitute \( T_r \) back to find \( a \)
Using \( T_r = 0.5 \) in Equation 3:
\[ a = \frac{50}{10 - 10(0.5)} \]
\[ a = \frac{50}{10 - 5} = \frac{50}{5} = 10 \, \text{m/s}^2 \]

### Step 6: Find the stopping distance for \( U_3 = 15 \, \text{m/s} \)
Using the same formula:
\[ S_3 = U_3 \cdot T_r + \frac{U_3^2}{2a} \]
\[ S_3 = 15 \cdot 0.5 + \frac{15^2}{2 \cdot 10} \]
\[ S_3 = 7.5 + \frac{225}{20} \]
\[ S_3 = 7.5 + 11.25 \]
\[ S_3 = 18.75 \, \text{m} \]

### Final Answer
The stopping distance when traveling at a speed of \( 15 \, \text{m/s} \) is \( 18.75 \, \text{m} \).