Q:

For most cars, the stopping distance from 30-0 is nearly exactly one-fourth of the distance from 60-0. As I understand, that makes sense, since energy increases as the square of speed.
However, the reverse is not true: 0-60 acceleration times generally are quite a bit less than four times 0-30 times. Any idea why?
Thanks very much.

- David (age 49)

Mill Valley, CA, USA

- David (age 49)

Mill Valley, CA, USA

A:

That's a nice little question David.

The basic reason is that for a constant de-acceleration rate, which maxes out when the tires start to skid on the pavement, you get a quadratic velocity-stopping distance relation. On the other hand the power going into acceleration is more or less a constant given by the engine. (The spinning of tires in drag races makes things worse.) For a given constant power input the acceleration rate is not a constant. The velocity only grows as the square root of time and the distance as the 3/2 power. That's the main difference. There are other subtle things like wind resistance, skidding, etc.

Lee

The basic reason is that for a constant de-acceleration rate, which maxes out when the tires start to skid on the pavement, you get a quadratic velocity-stopping distance relation. On the other hand the power going into acceleration is more or less a constant given by the engine. (The spinning of tires in drag races makes things worse.) For a given constant power input the acceleration rate is not a constant. The velocity only grows as the square root of time and the distance as the 3/2 power. That's the main difference. There are other subtle things like wind resistance, skidding, etc.

Lee

*(published on 07/11/2012)*