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 step by step, we will use the equations of motion. The problem involves two scenarios for a car that is braking.

### Given:
1. A car moving with speed \( v \) stops in distance \( d \).
2. The same car moving with speed \( 3v \) stops with half the retardation.

### Step 1: Analyze the first scenario
- Initial velocity \( u = v \)
- Final velocity \( v = 0 \)
- Distance \( s = d \)
- Let the retardation be \( a \).

Using the equation of motion:
\[ v^2 = u^2 + 2as \]
Substituting the known values:
\[ 0 = v^2 + 2(-a)d \]
This simplifies to:
\[ v^2 = 2ad \quad \text{(Equation 1)} \]

### Step 2: Analyze the second scenario
- Initial velocity \( u = 3v \)
- Final velocity \( v = 0 \)
- Distance \( s = d' \)
- The retardation is half, so it is \( \frac{a}{2} \).

Using the same equation of motion:
\[ v^2 = u^2 + 2as \]
Substituting the known values:
\[ 0 = (3v)^2 + 2\left(-\frac{a}{2}\right)d' \]
This simplifies to:
\[ 0 = 9v^2 - ad' \]
Rearranging gives:
\[ ad' = 9v^2 \quad \text{(Equation 2)} \]

### Step 3: Substitute Equation 1 into Equation 2
From Equation 1, we have:
\[ a = \frac{v^2}{2d} \]
Now substitute this value of \( a \) into Equation 2:
\[ \left(\frac{v^2}{2d}\right)d' = 9v^2 \]
Cancelling \( v^2 \) (assuming \( v \neq 0 \)):
\[ \frac{d'}{2d} = 9 \]
Multiplying both sides by \( 2d \):
\[ d' = 18d \]

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

### Final Answer
The distance \( d' \) is \( 18d \).

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