Re: braking distance

We will make the assumption that the acceleration provided by the tires on the road remains constant. Actually, there are two ways to compute the braking distance, which we will take one at a time.

First, the kinematic method. You are aware, I think, of the equation which relates the distance travelled to the quantities of acceleration and velocity:
x = 0.5 at² + v_ot + x_o
where a is the acceleration, t is the time interval, v_o is the starting velocity, and x_o is the starting distance. In this problem we will assign zero to the starting distance, so we get
x = 0.5 at² + v_ot (equation 1)
Another equation, which we need in order to remove time, t, from our first equation, is
v = at + v_o
and we want to arrive at v = 0, so we get
0 = at + v_o
from which we can compute
t = -v_o/a (equation 2)
Putting this value for t into equation 1, we get
x = -0.5 v_o²/a
and, since a < 0, we get
x = 0.5 v_o²/a

Second, the energy method. The original kinetic energy of the car does work against the road, by way of the tires, during the stop. Work is the force times the distance, and since force is the mass times acceleration, we get
work = max
where m is the mass of the car, a is the acceleration, and x is the stopping distance. Since the kinetic energy is
KE = 0.5 mv²
and since the beginning kinetic energy is totally "used up" doing work against the road, we can equate the kinetic energy and the work:
KE = work
0.5 mv² = max
Notice that the car's mass can be removed from this equation because it appears on both sides. We obtain
x = 0.5 v_o²/a

Using either method, the final equation is
x = 0.5 v_o²/a
Therefore, if the initial velocity is doubled then the stopping distance is 4 times as great.

John Link, MadSci Physicist