A car traveling at 34.8 m/s approaches the turnoff for a restaurant. If the driver slams on the brakes with an acceleration of -2.6 m/s2, what will be her stopping distance?
v = Vi + a t
x = Xi + Vi t + (1/2) a t^2
here
Vi = + 34.8 m/s
Xi = call it 0 at start of braking
a = -2.5 m/s^2
so
v = Vi + a t
0 = 34.8 - 2.5 t
so t = 34.8 / 2.5 seconds to stop
x = 0 + 34.8 t - (1/2)(2.5) t^2
or just x = average speed (half of max) times t