Question

Applications. The distance traveled by a freely falling body is a function of the elapsed time \(t:\) $$ f(t)=v_{0} t+\frac{1}{2} g t^{2} \quad \text { ft } $$ where \(v_{0}\) is the initial velocity and \(g\) is the acceleration due to gravity \(\left(32.2 \mathrm{ft} / \mathrm{s}^{2}\right)\) If \(v_{0}\) is \(55.0 \mathrm{ft} / \mathrm{s},\) find \(f(10.0), f(15.0),\) and \(f(20.0)\)

Short answer

The distances are f(10.0) = 2160.0 ft, f(15.0) = 4447.5 ft, and f(20.0) = 7540.0 ft.

Step-by-step solution

- Understand and Prepare the Function

Recognize that the function given describes the distance in feet (\text{ft}) traveled by a freely falling body given the time in seconds (\text{s}). The formula provided is $$f(t)=v_{0} t + \frac{1}{2} g t^{2}$$where \(v_{0}\) represents the initial velocity and \(g\) is the acceleration due to gravity. In this case, \(v_{0}=55.0 \text{ft/s}\) and \(g=32.2 \text{ft/s}^2\). The task is to calculate the distances at different times: \(f(10.0)\), \(f(15.0)\), and \(f(20.0)\).

- Calculate f(10.0)

To find the distance traveled at 10 seconds, substitute 10 for \(t\) in the equation: \[f(10.0) = 55.0 \times 10 + \frac{1}{2} \times 32.2 \times 10^2\]. Perform the calculations with each term separately: Initial velocity term is \(55.0 \times 10 = 550.0 \text{ft}\) and the acceleration term is \(\frac{1}{2} \times 32.2 \times 10^2 = 1610.0 \text{ft}\). Adding these two gives us the total distance: \(550.0 + 1610.0 = 2160.0 \text{ft}\).

- Calculate f(15.0)

To find the distance traveled at 15 seconds, substitute 15 for \(t\) in the equation: \[f(15.0) = 55.0 \times 15 + \frac{1}{2} \times 32.2 \times 15^2\]. Perform the calculations for each term: Initial velocity term is \(55.0 \times 15 = 825.0 \text{ft}\) and the acceleration term is \(\frac{1}{2} \times 32.2 \times 15^2 = 3622.5 \text{ft}\). Adding these gives us the total distance: \(825.0 + 3622.5 = 4447.5 \text{ft}\).

- Calculate f(20.0)

To find the distance traveled at 20 seconds, substitute 20 for \(t\) in the equation: \[f(20.0) = 55.0 \times 20 + \frac{1}{2} \times 32.2 \times 20^2\]. Perform the calculations for each term: Initial velocity term is \(55.0 \times 20 = 1100.0 \text{ft}\) and the acceleration term is \(\frac{1}{2} \times 32.2 \times 20^2 = 6440.0 \text{ft}\). Adding these gives us the total distance: \(1100.0 + 6440.0 = 7540.0 \text{ft}\).

Key concepts

Kinematic Equations

When analyzing the motion of objects, kinematic equations allow us to predict the future position of a moving body, starting from its initial position and taking into account its velocity and acceleration over time. These equations of motion are fundamental in physics and underpin many aspects of classical mechanics.

For a freely falling body, where acceleration is constant due to gravity, the kinematic equations become especially useful. The distance \(f(t)\) that the body travels over time \(t\) can be calculated using one such kinematic equation: \[ f(t)=v_{0} t+\frac{1}{2} g t^{2}\]In this formula, \(v_{0}\) represents the initial velocity, while \(g\) is the acceleration due to gravity. These equations are highly versatile, providing a quick method to solve problems related to motion with just a few known variables.

Acceleration Due to Gravity

One of the fundamental forces we constantly experience is gravity, which pulls objects towards the center of the Earth. The acceleration due to gravity is denoted by \(g\), and it is approximately \(9.8 \mathrm{m/s}^{2}\) or \(32.2 \mathrm{ft/s}^{2}\) depending on the unit system in use.

This constant acceleration is what gives weight to objects and causes freely falling bodies to accelerate downward, regardless of their mass. Understanding \(g\) is crucial for accurately predicting how an object will move when dropped or thrown near Earth's surface. The value might differ slightly based on altitude and geographical location, but for most homework problems, it is sufficient to use the standard average value provided.

Initial Velocity

When a body starts its motion, it may already have some velocity, known as the initial velocity, expressed as \(v_{0}\). This is the speed at which an object begins its journey before any external forces, such as gravity, begin to change its velocity.

In the context of a freely falling body, if an object is dropped, its initial velocity \(v_{0}\) is zero since it begins from rest. However, if the object is thrown downwards or upwards, \(v_{0}\) would be the initial speed in the direction of the throw. The direction is significant: if upwards, \(v_{0}\) is considered negative in the kinematic equations due to the opposite direction to gravity; if downwards, it is positive.