Asked by christine
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
Answers
Answered by
Henry
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.
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.
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