Stopping distance

“If you can read this, you’re too close.” Have you ever seen a bumper sticker that says this? It is used to discourage tailgating—the act of driving too close to the moving vehicle in front of you. How close is too close? The work–energy theorem holds the key to this practical physics question. Read the text aloud
Stopping distanceSuppose a car is traveling at 30 m/s (67 mph). The maximum force of traction between the road and a car is typically about 0.8mg, or 80% of the car’s weight. How much distance does the car need to stop? In a real situation there would be other factors, such as driver reaction time, but in this physics problem assume zero reaction time. The work done by the reaction force from braking decreases kinetic energy to zero. Read the text aloud Show Brake fade and glazing
Asked: distance d
Given: v = 30 m/s; F = 0.8mg
Relationships: W = Fd; Ek = ½mv2
Solution: Set the work done equal to the change in kinetic energy:
Fd = ½mv2 → 0.8mgd = ½mv2
d = v2/(2×0.8g) = (30 m/s)2/(2×0.8×9.8 m/s2) = 57 m (190 ft)
We can draw a very important point from the solution to the problem above: Stopping distance is proportional to the square of velocity, or d ∝ v2. Twice the speed means four times the stopping distance! This means that a car initially traveling at 60 mph needs four times the stopping distance compared to a car traveling 30 mph. And that’s not all. If you see a hazard ahead it takes time—your reaction time tr— to apply the brakes. Since distance equals the product of speed and time, this will add vi×tr to the amount we calculated earlier. At 30 m/s, each one-half second of reaction time adds another 15 m of stopping distance, or almost 50 ft. Read the text aloud Show How do antilock brakes help?
Stopping distance involves both reaction time and work done by friction.
When accounting for friction between tires and the road for a skidding vehicle, the force of sliding friction is μkmg, not mg (or 0.8mg as in the problem above). The equation for stopping distance is then d = v2/(2μkg). If the coefficient of kinetic friction μk is small, as is the case for a wet or icy surface, then the stopping distance becomes very large! Read the text aloud
Suppose a car has a speed of 20 m/s and suddenly brakes. The car then skids for a distance of 10 m before coming to a stop. How far would the car have skidded if it had been traveling at 60 m/s instead? Show
Assume that the coefficient of kinetic friction μk between a car’s tires and the roadway is 0.8 when dry and 0.4 when wet. Now imagine that a driver suddenly brakes, and her car skids 15 m on dry pavement. How far would the car have skidded if the pavement had been wet? (Assume that the initial speed is the same in both cases.) Show

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