A car moving with speed v on a straight road can be stopped with in distance d on applying brakes. If same car is moving with speed 3v and brakes provide half retardation, then car will stop after travelling distance

Options

A. 6d
B. 3d
C. 9d
D. 18d

Answer

To solve the problem, we will use the equations of motion, specifically the third equation of motion, which relates initial velocity, final velocity, acceleration, and distance.

### Step-by-Step Solution:

1. **Identify the known values:**
- Initial speed of the car when it can stop: \( v \)
- Stopping distance at speed \( v \): \( d \)
- New initial speed of the car: \( 3v \)
- New retardation (deceleration) when brakes are applied: \( \frac{A}{2} \) (where \( A \) is the original retardation).

2. **Use the third equation of motion:**
The third equation of motion is given by:
\[ v^2 = u^2 + 2a s \]
where:
- \( v \) = final velocity (0 when the car stops),
- \( u \) = initial velocity,
- \( a \) = acceleration (negative for deceleration),
- \( s \) = distance traveled.

3. **For the first scenario (speed \( v \)):**
- Initial velocity \( u = v \)
- Final velocity \( v = 0 \)
- Acceleration \( a = -A \)
- Distance \( s = d \)

Plugging into the equation:
\[ 0 = v^2 - 2Ad \]
Rearranging gives:
\[ v^2 = 2Ad \quad \text{(1)} \]

4. **For the second scenario (speed \( 3v \)):**
- Initial velocity \( u = 3v \)
- Final velocity \( v = 0 \)
- Acceleration \( a = -\frac{A}{2} \)
- Distance \( s = d' \) (the distance we want to find)

Plugging into the equation:
\[ 0 = (3v)^2 - 2\left(-\frac{A}{2}\right)d' \]
Simplifying gives:
\[ 0 = 9v^2 + Ad' \]
Rearranging gives:
\[ Ad' = 9v^2 \quad \text{(2)} \]

5. **Substituting equation (1) into equation (2):**
From equation (1), we know:
\[ A = \frac{v^2}{2d} \]
Substituting this into equation (2):
\[ \left(\frac{v^2}{2d}\right)d' = 9v^2 \]
Simplifying gives:
\[ d' = 18d \]

### Final Answer:
The car will stop after traveling a distance of \( 18d \).