Problem 31 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

During the construction of a high-rise building, a worker accidentally dropped his portable electric screwdriver from a height of \(400 \mathrm{ft}\). After \(t\) sec, the screwdriver had fallen a distance of \(s=16 t^{2} \mathrm{ft}\). a. How long did it take the screwdriver to reach the ground? b. What was the average velocity of the screwdriver between the time it was dropped and the time it hit the ground? c. What was the velocity of the screwdriver at the time it hit the ground?

Step-by-Step Solution

Answer

The screwdriver took \(5 \, \mathrm{sec}\) to reach the ground. The average velocity during the fall was \(80 \, \mathrm{ft/sec}\), and the final velocity when it hit the ground was \(160 \, \mathrm{ft/sec}\).

Step 1: a. Time to reach the ground

Given the equation for the distance fallen, \(s = 16t^2\). To find how long it took the screwdriver to reach the ground, we can set \(s = 400\) ft and solve for \(t\). \[400 = 16t^2\] Divide both sides by 16: \[25 = t^2\] Now, we can take the square root of both sides: \[t = 5 \, \mathrm{sec}\] So, the screwdriver took \(5 \, \mathrm{sec}\) to reach the ground.

Step 2: b. Average velocity

To find the average velocity, we can use the formula \(v_{avg} = \dfrac{s_f - s_i}{t_f - t_i}\). Since \(s_f = 400 \, \mathrm{ft}\), \(s_i = 0 \, \mathrm{ft}\), and \(t_f - t_i = 5 - 0 = 5 \, \mathrm{sec}\), we can calculate the average velocity as: \[v_{avg} = \frac{400-0}{5}\] \[v_{avg} = \frac{400}{5}\] \[v_{avg} = 80 \, \mathrm{ft/sec}\] The average velocity of the screwdriver during the fall was \(80 \, \mathrm{ft/sec}\).

Step 3: c. Final velocity

To find the final velocity at the moment the screwdriver hits the ground, we first need to determine the equation for velocity. We can do this by taking the derivative of the given distance equation with respect to time: \[s = 16t^2\] Derive both sides with respect to \(t\): \[v = \frac{ds}{dt} = 16(2t)\] \[v = 32t\] Now that we know the velocity equation, we can find the velocity at the time it hit the ground, i.e., when \(t = 5 \mathrm{sec}\): \[v = 32(5)\] \[v = 160 \, \mathrm{ft/sec}\] The final velocity of the screwdriver when it hit the ground was \(160 \, \mathrm{ft/sec}\).

Key Concepts

Understanding Average Velocity

Average velocity is one of the primary concepts in kinematics, and it's crucial for understanding motion over a period of time. Unlike speed, which tells us how fast an object is moving at a given moment, average velocity provides an overall rate of change of position for the entire journey of an object. In this exercise, the screwdriver was dropped from a height and fell to the ground in a straightforward and linear motion.

To calculate the average velocity, use the formula:

• \( v_{\text{avg}} = \frac{s_f - s_i}{t_f - t_i} \)

Here:

• \(s_f\) is the final position (400 ft).
• \(s_i\) is the initial position (0 ft).
• \(t_f - t_i\) is the total time interval (5 seconds).

Substituting these values gives us \(v_{\text{avg}} = \frac{400 - 0}{5} = 80\) ft/sec. Essentially, this means that over the entire fall, the screwdriver moved at an average rate of 80 feet per second towards the ground.

Analyzing average velocity helps students understand that because gravity uniformly accelerates objects, the average velocity is influenced by both the initial motionless state and the increasing speed as the object falls.

Exploring Derivatives in Physics

Derivatives serve as powerful tools in physics, especially when analyzing how quantities like velocity and acceleration change over time. They allow us to calculate instantaneous rates of change, which is pivotal when dealing with motion.

In this exercise, we start with a distance-time relationship given by the equation \(s = 16t^2\). The goal is to find the velocity of the falling object. To do this, we derive this equation with respect to time using basic principles of calculus:

• The derivative of \(s = 16t^2\) is \(v = \frac{ds}{dt} = 32t\).

This derivative, \(v = 32t\), provides the velocity as a function of time. By substituting different values of \(t\) into this function, we can find out how fast the object is moving at any particular moment during its fall.

For example, at precisely \(t = 5\) seconds (the time the screwdriver hit the ground), the velocity \(v = 32 \cdot 5 = 160\) ft/sec. This calculation shows how derivatives give us insight into how fundamental physics quantities such as velocity can be rigorously understood as time varies.

Calculating Falling Objects

Falling object problems are a staple in physics and help illustrate the influence of gravity on motion. When an object is dropped from a certain height, gravity acts as the sole force accelerating it downward, giving it both an increasing velocity and a well-defined distance-time relationship.

In the case of the dropped screwdriver, its motion was dictated by the formula \(s = 16t^2\). This equation implies that the distance the screwdriver falls is proportional to the square of the time \(t\). Initially given a height of 400 ft, our task was to ascertain the time needed for the screwdriver to reach the ground.

Applying basic algebra:

• Set \(s = 400\) to find \(t\).
• Solve \(400 = 16t^2\) by dividing by 16.
• This equation simplifies to \(25 = t^2\), giving \(t = 5\) seconds when solved.

This simple arithmetic demonstrates the methodology physicists use to calculate the time frame of a fall with just an initial height and the gravitational acceleration effect encapsulated in the equation. Such calculations lay the groundwork for more advanced explorations in kinematic motion and free-fall dynamics.