Problem 21 - Algebra A Combined Function

Question

Pairs of markings a set distance apart are made on highways so that police can detect drivers exceeding the speed limit. Over a fixed distance, the speed \(R\) varies inversely with the time \(T\). In one particular pair of markings, \(R\) is 45 mph when \(T\) is 6 seconds. Find the speed of a car that travels the given distance in 5 seconds.

Step-by-Step Solution

Answer

The speed of the car is 54 mph.

Step 1: Understand Relationships

First, recall that speed \(R\) varies inversely with time \(T\) over a fixed distance. This relationship can be expressed as \(R = \frac{k}{T}\), where \(k\) is a constant.

Step 2: Determine the Constant

Given that \(R = 45\) mph when \(T = 6\) seconds, we can find \(k\) by substituting these values into the inverse variation formula: \(45 = \frac{k}{6}\). Solving for \(k\) gives us \(k = 45 \times 6 = 270\).

Step 3: Find Unknown Speed

Now that we know \(k = 270\), we use the formula again to find the new speed when \(T = 5\) seconds: \(R = \frac{270}{5}\).

Step 4: Solve for Speed

Calculate \(R\) when \(T = 5\) seconds: \(R = \frac{270}{5} = 54\) mph. Hence, the car's speed is 54 mph.

Key Concepts

Speed Calculation

Calculating speed involves understanding the relationship between distance and time. Speed is defined as the distance traveled per unit of time. For this, we use the formula:
\[R = \frac{D}{T}\]
Where:

• \( R \) is the speed.
• \( D \) is the distance traveled.
• \( T \) is the time taken to cover that distance.

When calculating speed in inverse variation problems like the one in our exercise, we're given that speed varies inversely with time for a fixed distance.
This means that instead of knowing the distance, the focus is on the relationship between speed and time. The faster you go, the less time you spend to cover a certain distance, and vice versa. In the exercise, we calculate the speed using this inverse relationship where the formula to be used becomes \( R = \frac{k}{T} \), a classic inverse variation equation.

Algebraic Expressions

Algebraic expressions are crucial for solving inverse variation problems.
They allow us to express relationships between variables in a formulaic way, helping us to calculate unknowns by isolating them with different operations.
An algebraic expression like \( R = \frac{k}{T} \) shows how speed is calculated through the concept of inverse variation. Here's how it breaks down:

• The constant \( k \) holds the key to solving the problem by bridging the relationship between speed and time.
• The expression \( \frac{k}{T} \) implies that as time \( T \) increases, the resulting speed \( R \) decreases, and vice versa.

Using algebraic manipulation, you can calculate the constant once you've established an initial condition. After determining \( k \), plug in the new value for \( T \) to solve for the new speed. This step-by-step methodical manipulation of the algebraic expression unfolds the hidden values effortlessly.

Inverse Relationships

Inverse relationships play a crucial role in understanding how one variable affects another in mathematical relationships. In such relationships, when one variable increases, the other decreases in proportion, provided all other conditions remain the same.
In the case of speed and time over a fixed distance, we use:
\[R = \frac{k}{T}\]Understanding this formula is key to grasping inverse variation. When you realize that increasing speed reduces travel time and vice versa, it becomes easier to solve problems with inverse relationships.

• By establishing \( k \), the constant, you can calculate any unknown variable as long as you have one variable and the constant.
• It’s the cyclical nature of the inverse relationship that allows such predictive calculations.

Recognizing and using inverse relationships can simplify complex calculations by focusing on proportional changes, making tasks like speed estimation much more efficient.