Problem 54 - Precalculus: Functions and Graphs

Question

The formula occurs in the indicated application. Solve for the specified variable. \(s=\frac{1}{2} g t^{2}+v_{0 t}\) for \(v_{0} \quad\) (distance an object falls)

Step-by-Step Solution

Answer

The initial velocity \(v_{0}\) is \(v_{0} = \frac{s}{t} - \frac{1}{2} g t\).

Step 1: Understand the Given Formula

The given formula is a kinematic equation for the distance an object falls under the influence of gravity: \(s = \frac{1}{2} g t^2 + v_{0} t\). We are asked to solve this equation for the initial velocity \(v_{0}\).

Step 2: Identify the Terms

Identify the terms in the equation: \(s\) is the distance fallen, \(\frac{1}{2} g t^2\) represents the influence of gravity, and \(v_{0} t\) is the term involving the initial velocity \(v_{0}\). We want to isolate \(v_{0}\).

Step 3: Rearrange the Equation

To solve for \(v_{0}\), we need to isolate the \(v_{0} t\) term. Start by subtracting \(\frac{1}{2} g t^2\) from both sides of the equation:\[s - \frac{1}{2} g t^2 = v_{0} t\]

Step 4: Solve for v_{0}

To solve for \(v_{0}\), divide both sides of the equation by \(t\) (assuming \(t eq 0\)):\[v_{0} = \frac{s - \frac{1}{2} g t^2}{t}\]

Step 5: Final Expression for v_{0}

The final expression shows \(v_{0}\) in terms of \(s\), \(g\), and \(t\): \[v_{0} = \frac{s}{t} - \frac{1}{2} g t\].

Key Concepts

Kinematic Equations

When studying motion, particularly motion under constant acceleration, kinematic equations become indispensable tools. These equations describe how objects move. They can relate different aspects of motion such as velocity, acceleration, time, and displacement. Kinematic equations help us predict what an object will do in a given scenario based on what we already know.

• In the equation \(s = \frac{1}{2} g t^2 + v_{0} t\), each symbol stands for a specific variable involved in the object's motion.
• \(s\) is the displacement or distance fallen, \(g\) is the acceleration due to gravity, \(t\) is the time, and \(v_0\) is the initial velocity.
• The beauty of kinematic equations is in their ability to be rearranged to solve for any one variable given the others.

By understanding each term and the physics behind it, you can solve problems related to motion more easily.

Initial Velocity

Initial velocity, symbolized as \(v_0\), is the velocity of an object before it undergoes acceleration. Initial velocity is crucial because it sets the starting conditions for any motion problem. Determining \(v_0\) allows you to predict the future state of the object's motion.

• In our context, \(v_0\) is the speed of the object at the start of its fall.
• Finding \(v_0\) requires arranging the equations to isolate this variable, as seen when solving \( v_{0} = \frac{s}{t} - \frac{1}{2} g t \).
• This means \(v_0\) can depend on other observed values like the distance fallen \(s\), time \(t\), and the gravitational acceleration \(g\).

Understanding how to manipulate the equation to isolate \(v_0\) can sometimes be key to solving more complex physics problems.

Distance Fallen

Distance fallen \(s\) is another major aspect of kinematic equations. It refers to how far an object moves from its initial position. Decomposing the components affecting \(s\) can be essential in solving motion problems.

• The formula \(s = \frac{1}{2} g t^2 + v_{0} t\) represents the combination of two effects on distance.
• \(\frac{1}{2} g t^2\) is the distance due to gravitational free fall, while \(v_{0} t\) accounts for the distance traveled due to the initial velocity.
• Knowing \(s\) allows you to work backward with the given formula to solve for unknowns like \(v_0\).

The typical use of \(s\) in kinematic calculations emphasizes the importance of observing movement and measuring how far an object goes during its journey.