shuffleboard disk is accelerated at a constant rate from rest to a speed of 6.0 m/s over a 1.5 m distance by a player using a cue. At this point the disk loses contact with the cue and slows at a constant rate of 3.5 m/s2 until it stops. (a) How much time elapses from when the disk begins to accelerate until it stops? (b) What total distance does the disk travel

Answer 1

a. a = (V^2-Vo^2)/2d.
a = (36-0)/3 = 12 m/s^2.
T1 = (V-Vo)/a = (6-0)/12 = 0.50 s. To reach 6 m/s.

T2 = (V-Vo)/a = (0-6)/-3.5=1.71 s.To
stop.
T1 + T2 = 0.5 + 1.71 = 2.21 s. = Total
time.

b. d = d1 + (V^2-Vo)/2a
d = 1.5 + (0-6)/-7 = 2.36 m.