Problem 18 - A Graphical Approach to Precalculus with Limits

Question

Solve each problem analytically, and support your solution graphically. Rumning At 2: 00 P.M. a runner heads north on a highway, jogging at 10 miles per hour. At 2: 30 P.M. a driver heads north on the same highway to pick up the runner. If the car travels at 55 miles per hour, how long will it take the driver to catch the runner?

Step-by-Step Solution

Answer

The car will take approximately 6.67 minutes to catch the runner.

Step 1: Calculate the Runner's Head Start

The runner starts jogging at 2:00 PM, and the car leaves at 2:30 PM. During these 30 minutes, the runner is jogging at a speed of 10 miles per hour. Therefore, in half an hour, the runner covers a distance of \( 10 \text{ miles/hour} \times 0.5 \text{ hours} = 5 \text{ miles} \).

Step 2: Set Up the Relative Motion Equation

The car travels at 55 mph, and the runner travels at 10 mph. The car is catching up to the runner at a relative speed of \( 55 - 10 = 45 \text{ mph} \).

Step 3: Calculate the Time Taken to Catch Up

The distance the car needs to cover to catch the runner is the initial 5-mile head start of the runner. The time \( t \) it takes for the car to catch the runner can be found using the equation \( 45t = 5 \). Solving for \( t \), we get \( t = \frac{5}{45} = \frac{1}{9} \text{ hours} \).

Step 4: Convert Time to Minutes for Practical Understanding

Since \( \frac{1}{9} \text{ hours} \) is not very intuitive in terms of time, we convert it to minutes: \( \frac{1}{9} \times 60 \text{ minutes} = \frac{60}{9} = \frac{20}{3} \approx 6.67 \text{ minutes} \).

Key Concepts

Distance-Time Relationship

Understanding the distance-time relationship is crucial when dealing with problems involving relative motion. In our scenario, we have a runner and a driver. Each one moves at a different constant speed. The runner begins his journey before the driver and covers some initial distance. This initial distance is important because it creates a head start that the driver must overcome to catch up.

To visualize this, think of a graph with time on the x-axis and distance on the y-axis. The runner's motion is represented as a steady line starting earlier, gradually increasing. The driver starts later, but because the car moves faster, its line is steeper. Where the lines intersect is the point where the driver catches up with the runner.

In simple terms:

• The runner's head start is the distance covered while the car was stationary. This is calculated by multiplying his speed by the time he was alone on the road.
• The car will quickly close this distance because it travels at a higher speed.

Rate of Change

Rate of change is another fundamental concept in studying motion. It refers to how quickly one quantity changes in relation to another. In the case of the runner and driver, the rate of change can be understood as the speed at which the car is catching up with the runner.

The runner jogs at 10 mph, while the driver zooms in at 55 mph. The relative speed, or the rate at which the driver catches up, is found by subtracting the runner's speed from the driver's speed: \[ 55 ext{ mph} - 10 ext{ mph} = 45 ext{ mph}.\]

This relative speed is crucial as it allows us to determine how fast the gap between them closes. To solve problems like this, always:

• Identify the individual speeds of objects involved.
• Calculate their relative speed, which is the difference in their speeds when moving in the same direction.
• Use this relative speed to find out how long it takes for them to catch up or separate.

Speed and Velocity

Speed and velocity are key components of motion. While they might seem similar, they have distinct meanings.

**Speed** is a scalar quantity, meaning it only considers how fast something is moving regardless of direction. In our problem:

• The runner's speed is 10 mph.
• The car's speed is 55 mph.

**Velocity**, on the other hand, is a vector quantity, which means it includes both speed and direction. Both runner and driver head north, keeping their velocity positive and straightforward.

Applying these concepts:

• You determine the runner's distance by multiplying speed by time.
• The driver's velocity allows it to close the gap based on their faster speed relative to the runner.

Thus, understanding the subtle difference between speed and velocity helps in problems involving direction, and knowing their implications aids in calculating the outcomes of relative motion problems effectively.