Problem 52 - College Algebra

Question

For the following exercises, use the given information to answer the questions. The velocity \(v\) of a falling object varies directly to the time, \(t,\) of the fall. If after 2 seconds, the velocity of the object is 64 feet per second, what is the velocity after 5 seconds?

Step-by-Step Solution

Answer

The velocity after 5 seconds is 160 feet per second.

Step 1: Understand the Problem

We need to find the velocity of a falling object after 5 seconds, knowing that it varies directly with time. We are given that after 2 seconds, the velocity is 64 feet per second.

Step 2: Establish the Direct Variation Equation

Since velocity varies directly with time, we can model this relationship with the equation \( v = kt \), where \( k \) is the constant of variation.

Step 3: Find the Constant of Variation

To find \( k \), use the given information: when \( t = 2 \), \( v = 64 \). Substitute these values into the equation: \( 64 = k \, \times\, 2 \). Solve for \( k \) to get \( k = 32 \).

Step 4: Use the Constant to Find the New Velocity

Now that we know \( k = 32 \), use it to find the velocity after 5 seconds by substituting \( t = 5 \) into the equation: \( v = 32 \, \times\, 5 \).

Step 5: Calculate the Velocity at 5 Seconds

Perform the multiplication: \( v = 32 \, \times\, 5 = 160 \). Thus, the velocity after 5 seconds is 160 feet per second.

Key Concepts

Velocity and Time Relationship

In the world of physics, understanding the relationship between velocity and time can greatly clarify how objects behave during motion. Simply put, when velocity varies directly with time, it means they increase or decrease proportionally to one another.
Consider a scenario where an object is dropped, free-falling under the force of gravity. If no other forces are acting on it (like air resistance), the velocity at which it falls increases directly with the amount of time it has been falling.
In the exercise, this relationship is described by the statement: "The velocity of a falling object varies directly to the time of the fall." This is often expressed in the equation form as:

• \( v = kt \)

Here, \( v \) represents velocity, \( t \) stands for time, and \( k \) is a constant that relates the two variables. Changes in velocity are directly dependent on the amount of time that has passed, provided the conditions remain constant.

Solving Algebraic Equations

Solving algebraic equations is a fundamental skill in mathematics. It allows us to find unknown values using relationships expressed with variables and constants. In our case, the equation \( v = kt \) can be solved to find values not immediately obvious.
The problem requires you to find the velocity after 5 seconds. Fortunately, a two-step algebraic process can reveal the answer:

• Substitute the known values into the equation. For the initial condition, it was: \( 64 = k \times 2 \).
• Solve for the constant of variation, \( k \). Here, it involves simple arithmetic: divide both sides by 2 to get \( k = 32 \).

Once \( k \) is determined, the problem becomes straightforward. Substitute \( k \) and the new time value back into the equation to calculate the new velocity.
By methodically solving equations like this, we can predict values of variables under direct variation circumstances.

Constant of Variation

The constant of variation, often denoted as \( k \), is crucial when dealing with direct variation equations. It acts as the bridge connecting the two variables and ensures their proportional relationship.
In the direct variation equation, \( v = kt \), \( k \) remains constant as long as the conditions of the problem do not change. It captures the specific characteristic of the relationship under study. Depending on the problem, \( k \) can have a different value, but within a particular scenario, it stays the same.
To find \( k \), we essentially solve for it using known values. In the exercise, by knowing that after 2 seconds the velocity is 64 feet per second, we could determine \( k \) to be 32. This consistent value of \( 32 \) then helps us calculate the velocity for any other time period using the basic equation \( v = 32t \).
Understanding how to determine and apply the constant of variation effectively can simplify solving many real-life problems that follow direct variation patterns.