Question: A shuffle board disk is accelerated at a constant rate from rest to a speed of 6.0 m/sover a 1.8 mdistance by a player using a cue. At this point the disk loses contact with the cue and slows at a constant rate of 2.5 m/s²until 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?

During the acceleration, the initial velocity, $v_0=0\text{m/s}$

During the acceleration, the final velocity, $v=6.0\text{m/s}$

During acceleration the distance traveled, $x_1=1.8\text{m}$

During slows down, the acceleration, $a_2=-2.5{\text{m/s}}^2$

This problem is based on Newton’s kinematic equations which describe the motion of an object with constant acceleration. Using these equations the acceleration of the disc when it is being pushed by the cue can be found. Further, the time for which the disc decelerates and the distance it travels while coming to rest can be calculated.

The expression for the kinematic equations of motion are given as follows:

$v=v_0+at$ … (i)

$v^2={v^2}_0+2ax$ … (ii)

Here,$v_0$ is the initial velocity, $t$is the time, $a$is the acceleration and $x$is the displacement.

Using equation (ii), the acceleration during first part of motion is,

$$\begin{gathered} a_1=\frac{v^2-{v_0}^2}{2x_1} \\ =\frac{{\left(6.0\text{m/s}\right)}^2-\left(0\text{m/s}\right)}{2\times 1.8\text{m}} \\ =10{\text{m/s}}^2 \end{gathered}$$

From equation (i), the time elapses during first part of motion is,

$$\begin{gathered} t_1=\frac{v-v_0}{a_1} \\ =\frac{6.0\text{m/s-0m/s}}{10{\text{m/s}}^2} \\ =0.6\text{s} \end{gathered}$$

For second part of motion,$v=0\text{m/s}$ and $v_0=6.0\text{m/s}$

Again from equation (i), the time elapses during second part of motion is,

Now, total time elapses during the whole motion is,

$$\begin{gathered} t_2=\frac{v-v_0}{a_2} \\ =\frac{0\text{m/s-6.0m/s}}{-2.5{\text{m/s}}^2} \\ =2.4\text{s} \end{gathered}$$

Thus, total time elapses during the whole motion is 3.0 s.

Using equation (ii), the distance travelled during the second part of motion is,

$$\begin{gathered} {\text{x}}_2\text{=}\frac{v^2-{v^2}_0}{2a_2} \\ =\frac{{\left(0\text{m}/\text{s}\right)}^2-{\left(6.0\text{m/s}\right)}^2}{2\times \left(-2.5{\text{m/s}}^2\right)} \\ =7.2\text{m} \end{gathered}$$

The total distance traveled during the whole motion is,

$$\begin{gathered} x{\text{=x}}_1{\text{+x}}_2 \\ \text{=1.8m+7.2m} \\ =9.0\text{m} \end{gathered}$$

Thus, the total distance traveled by disk is 9.0 m.