F = m A
so
A = -.057/.228
v = Vi + A t
0 = 10 - (.057/.228)t
t = 2.28/.057
d = 10 t -(.057/.456) t^2
A hockey puck of mass 0.228 kg si sliding across level ice. There is a force of friction of 0.057 N acting opposite the direction of motion of the puck while it is sliding.
a) what is the acceleration of the puck while it is sliding?
b)if it initially has a speed of 10.0 m/sec, how long will it take the friction to stop the puck, and how far will it slide in this time?
2 answers
or for d just use average speed during stop
5 * t
5 * t