Problem 12 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

A bicyclist does a one-mile climb at a constant speed of 12 miles per hour followed by a one-mile descent at a constant speed of 30 miles per hour. (a) Sketch a graph of distance traveled as a function of time. Assume the cyclist starts at \(t=0\) minutes, and be sure to label the times at which he reaches the top and bottom of the hill. (b) What is his average speed for the two miles? Is this the same as the average of 12 mph and \(30 \mathrm{mph}\) ? Explain why or why not.

Step-by-Step Solution

Answer

The average speed for the two miles is approximately 17.14 mph, not the average of 12 mph and 30 mph. This is due to the fact that the speed rates weren't maintained for identical time periods.

Step 1: Finding Time

Firstly, determine the time it takes to travel uphill and downhill separately. The time \(t\) it takes to cover a certain distance \(d\) at a certain speed \(s\) is found by the formula \(t =d/s\). Apply this formula for each scenario:\nUphill: \(t = 1/12\) hours which is 5 minutes\nDownhill: \(t = 1/30\) hours which is 2 minutes

Step 2: Sketching the Graph

Now sketch a graph where time (in minutes) is represented on the x-axis and distance traveled (in miles) on the y-axis. From the origin (0,0), draw a line with a slope representing 12 mph up to (5,1) marking the time and distance of the uphill part of the route, then draw a line with a slope representing 30 mph from (5,1) to (7,2), marking the time and distance of the downhill part of the route.

Step 3: Calculating Average Speed

The total distance the cyclist traveled is 2 miles (1 mile uphill and 1 mile downhill), and the total time taken is 7 minutes (5 minutes uphill and 2 minutes downhill). The average speed is given by total distance divided by total time, or \(2/ (7/60) = 17.14\) miles per hour.

Step 4: Comparing and Explaining Average Speed

The average speed of the cyclist is not simply the average of his uphill and downhill speeds (12 mph and 30 mph), which would be 21 mph. This is because the time spent at each speed is not equal. The cyclist spends more time at the slower speed, which pulls down the average speed. The average speed is a weighted average, heavily influenced by the slower speed because of the longer time spent at this speed.

Key Concepts

Distance-Time Graph

When working with problems involving motion, a distance-time graph is a powerful visual tool. It helps us understand how distance changes over time. On this graph, time is typically on the horizontal axis, while distance is on the vertical axis.
In this exercise, the cyclist's journey is split into two segments: uphill and downhill. Starting from the origin (0,0), where the journey begins, draw a line that uses the cyclist's uphill speed of 12 mph. This section ends at the point (5,1), representing the end of the uphill climb, taking 5 minutes to cover 1 mile.
Next, change the slope to represent the downhill speed of 30 mph. Draw this line from (5,1) to (7,2), showing that the cyclist covers the next mile in 2 minutes. Each segment's slope corresponds to the cyclist's speed. Steeper slopes indicate higher speeds; hence, the steeper line during the descent.

Constant Speed

Constant speed means that the cyclist maintains a steady pace throughout each segment. For instance, while climbing, the cyclist moves at 12 mph without accelerating or decelerating. Understanding constant speed helps to simplify calculations related to motion.
When speed remains constant, distance can easily be computed using the formula:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]
This formula is straightforward because it assumes no variation in speed. It allows for easy sketching of linear sections on a distance-time graph. Constant speed is reflected by a straight line with a constant gradient on the distance-time graph for each segment.

Weighted Average

Calculating a cyclist's average speed over changing speeds involves a weighted average. This is key when the time spent on each speed varies significantly. A simple mean of the speeds won’t accurately reflect the situation when the time distribution is unequal.
Instead of averaging 12 mph and 30 mph directly, the cyclist's performance is better evaluated using weighted average, considering:

• Total time: 7 minutes (or \( \frac{7}{60} \) hours)
• Total distance: 2 miles

The weighted calculation gives:\[ \text{Average Speed} = \frac{2}{7/60} = 17.14 \, \text{mph} \]
This example showcases how longer durations at lower speeds decrease the overall average. Weighted averages are crucial in understanding net performance over varied conditions.

Distance and Time Calculations

Understanding the relationship between distance, time, and speed is fundamental to solving motion problems. Several calculations in this exercise are based on the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]
For the uphill journey, the cyclist travels 1 mile at 12 mph, thus taking \( 1/12 \) hours, which equals 5 minutes. For the downhill segment, the same distance at 30 mph results in \( 1/30 \) hours, or 2 minutes.
Precise time calculations enable plotting accurate distance-time graphs and computing average speeds. Efficient use of the distance-time-speed triangle (where you remember these relationships by covering what you need: \( \text{Dst} \) – Distance = Speed \( \times \) Time) simplifies these calculations. Getting a clear grasp of these foundational concepts is essential for problem-solving and understanding motion in physics.