It takes a minimum distance of 41.14 m to stop a car moving at 11.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 28.0 m/s.

Find the time it took for the car to stop at 11.0m/s

V = deltax/t

t = 41.14/11.0 = 3.74s

Now find at what rate it was decelerating, so find the acceleration during that interval of time.

vf = vi + at

-11.0m/s = a3.74s

a = -2.94m/s^2

The acceleration is negative because is pulling the car towards its opposite direction to make it stop.

Now find how much time it would take for the car to stop at 28.0m/s but with the same acceleration, the car is the same so its acceleration to stop the car will remain the same.

vf = vi + at

0 = 28.0 - 2.94t

t = 9.52

Once the time is obtained, you can find the final position, xf, by plugging the time acceleration and velocity values.

xf = 0 + (28m/s)(9.52s) + 1/2(-2.94)(9.52s)^2

xf = 266.6m - 133.23m = 133m

a and b sometimes have a positive value