Problem 106 - Fundamentals of Physics

Question

A shuffleboard disk is accelerated at a constant rate from rest to a speed of \(6.0 \mathrm{~m} / \mathrm{s}\) over a \(1.8 \mathrm{~m}\) distance by a player using a cue. At this point the disk loses contact with the cue and slows at a constant rate of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\) 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?

Step-by-Step Solution

Answer

3.0 s; 9.0 m.

Step 1: Determine Initial Acceleration

Since the shuffleboard disk starts from rest and accelerates to a speed of \(6.0 \, \mathrm{m/s}\) over a distance of \(1.8 \, \mathrm{m}\), we need to find the acceleration. Use the kinematic equation \(v^2 = u^2 + 2as\). Here, \(u = 0\), \(v = 6.0 \, \mathrm{m/s}\), and \(s = 1.8 \, \mathrm{m}\). Substitute these values to find \(a\):\[6.0^2 = 0 + 2a \times 1.8\]\[36.0 = 3.6a\]Therefore, \(a = \frac{36.0}{3.6} = 10 \, \mathrm{m/s}^2\).

Step 2: Calculate Time of Acceleration

We now calculate the time taken to reach \(6.0 \, \mathrm{m/s}\) using the formula \(v = u + at\). Here, \(u = 0\), \(v = 6.0 \, \mathrm{m/s}\), and \(a = 10 \, \mathrm{m/s}^2\). Solve for \(t\):\[6.0 = 0 + 10t\]\[t = \frac{6.0}{10} = 0.6 \, \mathrm{s}\].

Step 3: Calculate Time to Stop After Losing Contact

After the disk loses contact with the cue, it slows at \(2.5 \, \mathrm{m/s}^2\) until it stops. Use the equation \(v = u + at\) to find the time to stop. Here, \(u = 6.0 \, \mathrm{m/s}\), \(v = 0\), and \(a = -2.5 \, \mathrm{m/s}^2\). Solve for \(t\):\[0 = 6.0 - 2.5t\]\[2.5t = 6.0\]\[t = \frac{6.0}{2.5} = 2.4 \, \mathrm{s}\].

Step 4: Total Time Elapsed

Add the time of acceleration to the time taken to stop to get the total time:\[\text{Total time} = 0.6 + 2.4 = 3.0 \, \mathrm{s}\].

Step 5: Calculate Distance After Losing Contact

The distance traveled while the disk slows to a stop can be found using the formula \(s = ut + \frac{1}{2}at^2\). Here, \(u = 6.0 \, \mathrm{m/s}\), \(a = -2.5 \, \mathrm{m/s}^2\), and \(t = 2.4 \, \mathrm{s}\):\[s = 6.0 \times 2.4 + \frac{1}{2}(-2.5)(2.4)^2\]\[s = 14.4 - 7.2\]\[s = 7.2 \, \mathrm{m}\].

Step 6: Total Distance Traveled

To find the total distance traveled, add the distance covered during acceleration to the distance after losing contact:\[\text{Total distance} = 1.8 + 7.2 = 9.0 \, \mathrm{m}\].

Key Concepts

Constant Acceleration

When we talk about constant acceleration, we're dealing with a situation where the rate of change of velocity remains the same over time. This means that, regardless of how much time passes, the acceleration value doesn't change. In the exercise with the shuffleboard disk, we see constant acceleration occurring in two phases. Initially, when the disk is pushed, it accelerates at a constant rate of 10 m/s² until it reaches a speed of 6 m/s. Constant acceleration is crucial because it allows us to use specific kinematic equations to predict motion accurately.

Here are some key points about constant acceleration:

• Acceleration remains unchanged, which simplifies calculations and predictions.
• Enables the use of specific mathematical formulas such as the kinematic equations.
• Commonly experienced in real-world scenarios, like vehicles accelerating or slowing down at steady rates.

Understanding constant acceleration is foundational in kinematics and allows us to model various real-life motions mathematically.

Kinematic Equations

Kinematic equations are formulas that describe the motion of objects under constant acceleration. These equations interrelate five key variables:

• Initial velocity (\(u\))
• Final velocity (\(v\))
• Acceleration (\(a\))
• Time (\(t\))
• Displacement (\(s\))

Kinematic equations were the backbone of solving the shuffleboard disk problem. Here’s a quick rundown of the equations:

• \(v = u + at\), used to find final velocity or time.
• \(s = ut + \frac{1}{2}at^2\), helpful for calculating the displacement.
• \(v^2 = u^2 + 2as\), to find final velocity when you know acceleration, initial velocity, and displacement.

In the shuffleboard problem, these equations were used to derive the acceleration, time elapsed, and the distance traveled during different phases of the disk's motion. They provide a reliable way to predict motion when acceleration is constant, as in this exercise.

Motion Analysis

Motion analysis involves breaking down the movement of an object and understanding how different forces affect it over time. In the context of the shuffleboard disk, motion analysis was used to examine two distinct phases: acceleration with the cue and deceleration after losing contact.

Initially, the disk undergoes acceleration due to the force exerted by a player, gaining speed up to 6 m/s over a distance of 1.8 meters. Once the disk loses contact, it transitions to a deceleration phase, slowing down until it comes to a complete stop. The constant deceleration was given as -2.5 m/s². By analyzing these phases through the lens of kinematic principles, we determine both the total time taken and the full journey length the disk completes from start to stop.

• Motion analysis helps in assessing each part of the motion in detail.
• Breaks down complex movement into manageable steps.
• Relies on understanding the influence of forces and resisting factors, like friction or drag.

Breaking down the shuffleboard disk's movement using motion analysis allows for a deeper understanding of how objects behave under constant acceleration conditions.