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A shuffleboard disk is accelerated at a constant rate from rest to a speed of 4.8 m/s over a 1.9 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 2.2 m/s$^2$ 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?

So, I have vo = 0, vf = 4.8 = (vo + vf) *t/2 = 9.6.

Then, x= 1/2 (vf+vo)*t(2.2) = 10.56.

It says the answers are wrong, though.

user366783
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2 Answers2

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You have two intervals of constant acceleration: the acceleration part, and the deceleration part.

The questions are asking you about both parts taken together. So,

  • Calculate the time over which the first part happens.
  • Calculate the time and distance over which the second part happens.
  • Add the times and distances for the two parts, and you're done.

Can you take it from here?

Some more hints:

$$x = x_0 + v_0t + at^2/2$$ $$v = v_0 + at$$

To calculate the time for the acceleration part, start with what you know. You know $a$, and you know $x$. (Both $x_0$ and $v_0$ are zero.) The first equation then becomes $t = \sqrt{2x/a}$.

John
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A slightly different approach that should lead to the same results:

  1. Compute the initial acceleration via conservation of energy: $\frac12v_{max}^2=a_{push}s$, where $a_{push}$ is the acceleration imparted by the player and $s$ the distance through which the puck is accelerated.
  2. Compute the distance it takes for the puck to come to a stop, again via conservation of energy.
  3. Use the fact the the accelerations are constant to find the times for acceleration and deceleration: $\Delta t=\Delta v/a$.
amd
  • 53,693