A car, moving with a speed of `50 km//hr`, can be stopped by brakes after at least `6 m`. If the same car is moving at a speed of `100 km//hr`, the minimum stopping distance is

Options

A. 6 m
B. 12 m
C. 18 m
D. 24 m

Explanation

To solve the problem of finding the minimum stopping distance of a car moving at a speed of `100 km/hr`, we can follow these steps:

### Step 1: Convert the speed from km/hr to m/s
The initial speed of the car when moving at `100 km/hr` needs to be converted to meters per second (m/s). The conversion factor is:
\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]
Thus,
\[ u = 100 \times \frac{5}{18} = \frac{500}{18} \text{ m/s} \approx 27.78 \text{ m/s} \]

### Step 2: Determine the deceleration (retardation)
From the first part of the problem, we know that when the car is moving at `50 km/hr`, it stops after `6 m`. We can use the third equation of motion to find the deceleration.

Using the equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v = 0 \) (final velocity when the car stops)
- \( u = \frac{250}{18} \text{ m/s} \) (initial speed at `50 km/hr`)
- \( s = 6 \text{ m} \) (stopping distance)

Substituting the values:
\[ 0 = \left(\frac{250}{18}\right)^2 + 2a(6) \]
This simplifies to:
\[ 0 = \frac{62500}{324} + 12a \]
Rearranging gives:
\[ 12a = -\frac{62500}{324} \]
Thus,
\[ a = -\frac{62500}{324 \times 12} \approx -16.07 \text{ m/s}^2 \]

### Step 3: Calculate the stopping distance for `100 km/hr`
Now, we can use the same equation of motion to find the stopping distance when the car is moving at `100 km/hr`.

Using:
\[ v^2 = u^2 + 2as \]
where:
- \( v = 0 \)
- \( u = 27.78 \text{ m/s} \) (from Step 1)
- \( a = -16.07 \text{ m/s}^2 \)

Substituting the values:
\[ 0 = (27.78)^2 + 2(-16.07)s \]
This simplifies to:
\[ 0 = 771.68 - 32.14s \]
Rearranging gives:
\[ 32.14s = 771.68 \]
Thus,
\[ s = \frac{771.68}{32.14} \approx 24.00 \text{ m} \]

### Conclusion
The minimum stopping distance for the car moving at `100 km/hr` is approximately `24 m`.

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