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

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.

To answer these questions, we need to break down the motion of the shuffleboard disk into two parts: the acceleration phase and the deceleration phase.

(a) Let's start with the acceleration phase. We are given that the disk is accelerated at a constant rate from rest to a speed of 6.0 m/s over a distance of 1.5 m. We can use the equation:

v = u + at

where:
- v is the final velocity (6.0 m/s),
- u is the initial velocity (0 m/s, as the disk starts from rest),
- a is the acceleration (which we need to find), and
- t is the time taken.

Rearranging the equation to solve for acceleration, we have:

a = (v - u) / t

Substituting the given values, we have:

a = (6.0 m/s - 0 m/s) / t
a = 6.0 m/s / t

Now, let's calculate the acceleration:

a = 6.0 m/s / t

Moving on to the deceleration phase, we know that the disk slows down at a constant rate of 3.5 m/s^2. The final velocity here is 0 m/s as the disk comes to a stop. Using the equation v = u + at once again, we have:

0 = 6.0 m/s + (-3.5 m/s^2) * t

Simplifying this equation, we find:

3.5 m/s^2 * t = 6.0 m/s

Now, we can solve for time in the deceleration phase:

t = 6.0 m/s / 3.5 m/s^2
t ≈ 1.71 s

So, the time taken for the disk to stop during the deceleration phase is approximately 1.71 seconds.

To find the total time, we add the time taken during the acceleration and deceleration phases:

Total time = time during acceleration + time during deceleration
Total time = t + t
Total time = 1.71 s + 1.71 s
Total time ≈ 3.42 s

Therefore, the total time elapsed from when the disk begins to accelerate until it stops is approximately 3.42 seconds.

(b) To find the total distance traveled by the disk, we need to sum up the distances covered during the acceleration and deceleration phases.

Distance during acceleration = (initial velocity * time) + (0.5 * acceleration * time^2)
Distance during acceleration = (0 m/s * t) + (0.5 * a * t^2)

Distance during deceleration = (initial velocity * time) + (0.5 * deceleration * time^2)
Distance during deceleration = (6.0 m/s * t) + (0.5 * -3.5 m/s^2 * t^2)

Now, let's calculate the distances:

Distance during acceleration = (0 m/s * 1.71 s) + (0.5 * a * (1.71 s)^2)

Distance during deceleration = (6.0 m/s * 1.71 s) + (0.5 * -3.5 m/s^2 * (1.71 s)^2)

Adding these distances together will give us the total distance traveled by the shuffleboard disk.

Total distance = Distance during acceleration + Distance during deceleration

Please provide the value of 'a' (acceleration during the acceleration phase) to continue the calculation.