Question

The minimum distance necessary for a car to brake to a stop from a speed of \(100.0 \mathrm{~km} / \mathrm{h}\) is \(40.00 \mathrm{~m}\) on a dry pavement. What is the minimum distance necessary for this car to brake to a stop from a speed of \(130.0 \mathrm{~km} / \mathrm{h}\) on dry pavement?

Short answer

Answer: The minimum distance necessary for the car to stop from a speed of 130 km/h on dry pavement is approximately 67.2 meters.

Step-by-step solution

Set up a proportion

We have the initial conditions (\(v_1 = 100 \mathrm{~km/h}\) and \(d_1 = 40 \mathrm{~m}\)) and the new speed (\(v_2 = 130 \mathrm{~km/h}\)). We can set up a proportion using the formula for braking distance: \(\frac{d_1}{v_1^2} = \frac{d_2}{v_2^2}\)

Convert speeds to meters per second

Convert the speeds from km/h to m/s by multiplying by the factor \(\frac{1000 \mathrm{~m}}{3600 \mathrm{~s}}\). \(v_1 = 100.0 \mathrm{~km/h} \cdot \frac{1000 \mathrm{~m}}{3600 \mathrm{~s}} = 27.8 \mathrm{~m/s}\) \(v_2 = 130.0 \mathrm{~km/h} \cdot \frac{1000 \mathrm{~m}}{3600 \mathrm{~s}} = 36.1 \mathrm{~m/s}\)

Calculate the new braking distance

Substitute the known values into the proportion and solve for the unknown variable \(d_2\): \(\frac{40 \mathrm{~m}}{(27.8 \mathrm{~m/s})^2} = \frac{d_2}{(36.1 \mathrm{~m/s})^2}\) \(d_2 = \frac{(36.1 \mathrm{~m/s})^2}{(27.8 \mathrm{~m/s})^2} \cdot 40 \mathrm{~m} = 67.2 \mathrm{~m}\) The minimum distance necessary for the car to brake to a stop from a speed of 130 km/h on dry pavement is approximately 67.2 meters.