Problem 52 - College Algebra with Corequisite Support

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 ft/s.

Step 1: Understand Direct Variation

When we say that velocity \(v\) varies directly with time \(t\), it means that \(v = kt\), where \(k\) is a constant. We need to find \(k\) to determine the velocity at any time \(t\).

Step 2: Find the Constant of Variation

We are given that after 2 seconds, the velocity is 64 feet per second. Substitute these values into the equation \(v = kt\) to find \(k\). So, \(64 = k \times 2\). To solve for \(k\), divide both sides by 2, giving \(k = 32\).

Step 3: Calculate the Velocity After 5 Seconds

Now that we have \(k = 32\), we can use the equation \(v = 32t\). Substitute \(t = 5\) to find the velocity after 5 seconds: \(v = 32 \times 5\).

Step 4: Simplify and Conclude

Perform the multiplication: \(v = 160\). So, after 5 seconds, the velocity of the object is 160 feet per second.

Key Concepts

Constant of Variation

In direct variation, a relationship between two variables exists such that if one variable changes, the other changes in proportion to the first. This is described by the equation \( v = kt \), where \( k \) is the **constant of variation**. This constant remains the same as long as the set conditions of the variation do not change.

When tackling a problem involving direct variation, the first step is usually to determine the constant of variation. In our exercise, we found that \( v = 64 \) feet per second when \( t = 2 \) seconds. Substituting these values into the equation, we have \( 64 = 2k \). Solving for \( k \) gives us \( k = 32 \).

Understanding the constant of variation helps us predict how one quantity changes in response to another. Consider this a scaling factor: every increase in time results in an increase in velocity, magnified by this constant factor of \( 32 \).

Velocity

Velocity is a measure of the speed of an object in a particular direction. In our scenario, velocity is directly proportional to the time the object has been falling. The longer the time, the greater the velocity.

The relation can be recalled by the equation \( v = kt \), where \( v \) represents velocity and \( k \) the constant of variation. This formula helps us calculate the velocity at any given time, provided the initial conditions and the constant are known.

In the problem, once the constant \( k = 32 \) was determined, we could calculate the velocity after 5 seconds using \( v = 32t \). Substituting \( t = 5 \), we get \( v = 160 \) feet per second. Hence, velocity helps us understand how fast an object is moving as a function of time.

Time

Time, denoted here by \( t \), is a crucial factor influencing directly varying quantities like velocity. In direct variation problems, time often acts as an independent variable; it initiates changes in other dependent variables.

In the given exercise, we use time to determine how velocity builds up over a period. When \( t = 2 \) seconds, the velocity was found to be \( 64 \) feet per second. Using this information to find the constant of variation, \( k \), allows us to predict velocities at different times.

After determining \( k \), calculating velocity at any future time is straightforward: just insert the desired time into the equation \( v = kt \). Here, using \( t = 5 \), we calculated the resulting velocity and found it to be \( 160 \) feet per second. Thus, time serves as the essential backdrop against which direct variation unfolds, linking it firmly with velocity.