A car has speed of 40km/h. on applying brakes it stops after 15m. If its speed was 80 `km h^(-1)` it would have stopped after

Options

A. 15M
B. 30M
C. 45M
D. 60M

Explanation

To solve the problem step by step, we will use the equations of motion and the relationship between speed, acceleration, and distance.

### Step 1: Identify the given data
- Initial speed of the car (u₁) = 40 km/h
- Stopping distance (s₁) = 15 m
- New initial speed (u₂) = 80 km/h

### Step 2: Convert speeds from km/h to m/s
To use the equations of motion, we need to convert the speeds from km/h to m/s:
- u₁ = 40 km/h = (40 * 1000 m) / (3600 s) = 11.11 m/s
- u₂ = 80 km/h = (80 * 1000 m) / (3600 s) = 22.22 m/s

### Step 3: Use the equation of motion
Using the equation of motion:
\[ v^2 = u^2 + 2as \]
Since the car comes to a stop, the final velocity (v) is 0. Thus, we can rearrange the equation to:
\[ 0 = u^2 + 2as \]
This can be rewritten as:
\[ u^2 = -2as \]
Since acceleration (a) is negative when the car is decelerating, we can express it as:
\[ a = -\frac{u^2}{2s} \]

### Step 4: Calculate acceleration for the first case
Using the first case data:
\[ a = -\frac{(11.11)^2}{2 \times 15} \]
Calculating this gives:
\[ a = -\frac{123.456}{30} = -4.1152 \, \text{m/s}^2 \]

### Step 5: Calculate stopping distance for the second case
Now, we will use the same formula for the second speed (u₂ = 22.22 m/s):
Using the same equation:
\[ s_2 = \frac{u_2^2}{-2a} \]
Substituting the value of acceleration calculated from the first case:
\[ s_2 = \frac{(22.22)^2}{-2 \times -4.1152} \]
Calculating this gives:
\[ s_2 = \frac{493.7284}{8.2304} \approx 60 \, \text{m} \]

### Final Answer
If the speed was 80 km/h, the car would have stopped after approximately **60 meters**.