Problem 33 - College Physics

Question

Starting from rest, a boulder rolls down a hill with constant acceleration and travels \(2.00 \mathrm{~m}\) during the first second. (a) How far does it travel during the second second? (b) How fast is it moving at the end of the first second? at the end of the second second?

Step-by-Step Solution

Answer

(a) 6 m; (b) 4 m/s at the end of the first second, 8 m/s at the end of the second second.

Step 1: Identify Known Variables

We know the following variables from the problem: initial velocity (\(u = 0\) m/s as it starts from rest), distance traveled in the first second (\(s_1 = 2.00\) m), and time of travel for this distance (\(t_1 = 1\) second).

Step 2: Use Equation of Motion for Acceleration

Using the equation of motion \(s = ut + \frac{1}{2}at^2\), we can find the acceleration \(a\). Since the initial velocity \(u = 0\), the equation simplifies to \(s_1 = \frac{1}{2}a(1)^2\). Substituting \(s_1 = 2\) m, we get \(2 = \frac{1}{2}a\). Solving for \(a\), we get \(a = 4\) m/s².

Step 3: Calculate Distance during Second Second

To find the distance during the second second, we first use the equation for distance \(s = ut + \frac{1}{2}at^2\) for the total 2 seconds, then subtract the distance of the first second from it. \(s_2 = ut + \frac{1}{2}a(2)^2 = 0 + \frac{1}{2} \times 4 \times 4 = 8\) m. During the first two seconds, it travels 8 m in total. Subtracting the first second's distance: \(s_{second} = 8 - 2 = 6\) m.

Step 4: Find Speed at the End of the First Second

The speed at the end of the first second can be found using \(v = u + at\), where \(u = 0\), \(a = 4\) m/s², and \(t = 1\) s. Thus, \(v_1 = 0 + 4 \times 1 = 4\) m/s.

Step 5: Find Speed at the End of the Second Second

Using the same formula \(v = u + at\), with \(t = 2\) seconds now, we find \(v_2 = 0 + 4 \times 2 = 8\) m/s.

Key Concepts

Equations of Motion

Equations of motion are fundamental formulas used to describe the mathematical relationship between velocity, acceleration, time, and displacement. When dealing with constant acceleration and motion starting from rest, these equations become especially handy.
Let's focus on the most crucial one for this scenario:

• Displacement Formula: \(s = ut + \frac{1}{2}at^2\)
  • \(s\) stands for the displacement, \(u\) is initial velocity, \(a\) is the constant acceleration, and \(t\) is the time over which the motion occurs.
  • This particular equation helps you calculate how far an object has traveled when it has a constant acceleration.

With the initial velocity \(u = 0\) (as the boulder starts from rest), the equation simplifies to \(s = \frac{1}{2}at^2\). Thus, with known values of \(s\) and \(t\), we can calculate the acceleration \(a\) if it's unknown. Each change in these variables is linearly related, giving us clear insights into motion analysis.

Distance Calculation

Understanding how to calculate distance in physics, especially under constant acceleration, can enable predictions about an object's future position.
The task of determining how far an object travels during a specific period can be solved using the equation of motion. For a boulder rolling down a hill with constant acceleration, if we need the distance during the boulder's second second of travel:

• First, calculate the total distance for the two seconds using the formula: \(s = ut + \frac{1}{2}at^2\).
• Next, substitute \(t\) with 2 seconds to find the entire distance traveled and then subtract the distance covered during the first second.
• For example: Total distance in 2 seconds is 8 meters; minus 2 meters from the first second results in 6 meters during the second second.

By following this method, it becomes straightforward to pinpoint any travel segment accurately, reinforcing the idea that physics relies heavily on systematic problem-solving.

Velocity Calculation

Velocity calculation involves determining an object's speed and direction at any given point in time. With constant acceleration, this becomes a smooth process thanks to specific formulas.
To find the speed of the boulder at the end of each second, the equation \(v = u + at\) is used:

• First Second:
  Considering that the initial velocity \(u = 0\) and acceleration \(a = 4\) m/s², after 1 second: \(v = 0 + 4 \times 1 = 4\) m/s.
• Second Second:
  Again, taking \(a = 4\) m/s², at the end of the second second (\(t = 2\)): \(v = 0 + 4 \times 2 = 8\) m/s.

This method allows easy understanding of how increasing intervals of time affect speed in the context of consistent acceleration. Therefore, these calculations not only tell us about speed changes but also help us better anticipate future speed at any given time point.