(a) If a car traveling 50 mph has a stopping distance of 150 ft, find the stopping distance of a car that is traveling 70 mph.
(b) If it takes 6 ft for a car to stop, how fast was it traveling before the brakes were applied?
(a) Using the formula for stopping distance: stopping distance = (initial velocity)^2 / (2 * μ * g)
where
μ = coefficient of friction between the tires and the road
g = acceleration due to gravity
Initial velocity = 50 mph = 73.33 ft/s
Stopping distance = 150 ft
Let stopping distance be S1 for the car traveling at 50 mph.
Therefore,
150 = (73.33)^2 / (2 * μ * 9.8)
μ = (73.33)^2 / (2 * 9.8 * 150)
μ ≈ 0.6
Now, let's find the stopping distance S2 for a car traveling at 70 mph (Initial velocity = 70 mph = 102.67 ft/s):
S2 = (102.67)^2 / (2 * 0.6 * 9.8)
S2 ≈ 366 ft
Therefore, the stopping distance for a car traveling at 70 mph is approximately 366 ft.
(b) Using the same formula for stopping distance:
Let stopping distance be S for the car before applying brakes.
Therefore,
6 = (initial velocity)^2 / (2 * μ * 9.8)
μ = (initial velocity)^2 / (2 * 9.8 * 6)
μ ≈ (initial velocity)^2 / 117.6
As the car was already traveling before applying the brakes, the initial velocity is the velocity before the brakes were applied.
Therefore, initial velocity ≈ sqrt(117.6 * μ)
Plugging in the value of μ ≈ 0.6, we get:
Initial velocity ≈ sqrt(117.6 * 0.6) ≈ 8.16 ft/s
Therefore, the car was traveling at approximately 8.16 ft/s before the brakes were applied. Converting this to mph:
8.16 ft/s * (3600 sec/hr) / (5280 ft/mile) ≈ 5.57 mph
So, the car was traveling at approximately 5.57 mph before the brakes were applied.