91Ó°ÊÓ

To solve this problem, let's first understand the given data: - At \(50 \mathrm{~km} / \mathrm{hr}\), the car can be stopped in a minimum distance of \(6 \mathrm{~m}\). - We need to find the minimum stopping distance at a speed of \(100\mathrm{~km} / \mathrm{hr}\). Let's assume the stopping distance (\(S\)) is proportional to the square of the speed (\(V^2\)). We want to find if there is a constant (\(k\)) such that this equation is true: \(S = kV^2\)

In order to find the constant \(k\), we need to plug in the initial values: \(6 \mathrm{~m} = k (50\mathrm{~km} / \mathrm{hr})^2\) However, the speed is given in \(\mathrm{km/hr}\), and the stopping distance is given in meters. To avoid any inconsistency, we need to convert the speed from \(\mathrm{km/hr}\) to \(\mathrm{m/s}\). To convert from \(\mathrm{km/hr}\) to \(\mathrm{m/s}\), we can use the following formula: \(1\mathrm{~km} / \mathrm{hr} = \frac{1000}{3600} \mathrm{~m} / \mathrm{s}\) So, the speed at \(50\mathrm{~km} / \mathrm{hr}\) in \(\mathrm{m/s}\) is: \(\frac{50 * 1000}{3600} = 13.88\mathrm{~m/s}\) Now, we can rewrite our initial equation as: \(6 = k(13.88)^2\) To solve for \(k\), we can divide both sides by \((13.88)^2\): \(k = \frac{6}{(13.88)^2} = 0.031\)

Now that we have the value of \(k\), we can use it to determine the stopping distance (\(S\)) at the new velocity of \(100\mathrm{~km} / \mathrm{hr}\): First, let's convert the speed to \(\mathrm{m/s}\): \(\frac{100 * 1000}{3600} = 27.77\mathrm{~m/s}\) Now, substitute the known values into our previously derived equation: \(S = 0.031 * (27.77)^2 = 23.96\) The stopping distance at \(100\mathrm{~km} / \mathrm{hr}\) is approximately \(24\mathrm{~m}\).

Based on our calculations, the minimum stopping distance at a speed of \(100\mathrm{~km} / \mathrm{hr}\) is approximately \(24\mathrm{~m}\). Therefore, the correct option is: (C) \(24 \mathrm{~m}\)