Vi=8.55m/s
Ff=Ukmg=0.45m(9.8m/s2)Calculate the acceleration (which will be negative since it's due to the force of friction):
a=Fma=(0.45)(9.8m/s2)mma=4.41m/s2
Time to stop:
t=8.55m/s4.41m/s2=1.94s
Now see how far they go in 1.94s with an acceleration of -4.41m/s^2:
x=V0t+12at2x=(8.55m/s)(1.94s)+12(−4.41m/s2)(1.94s)2=8.29m
