Problem 36 - College Physics

Question

A student was 18 in. long when she was born. She is now \(5 \mathrm{ft} 6\) in. tall and 20 years old. How many centimeters a year did she grow on average?

Step-by-Step Solution

Answer

She grew on average 6.096 cm per year.

Step 1: Convert height to inches

Convert the current height of the student from feet and inches to inches only. The student is 5 feet 6 inches tall. We know that 1 foot equals 12 inches, so we calculate: \(5 \times 12 + 6 = 66\) inches.

Step 2: Find total growth in inches

Subtract the height at birth from the current height to find the total growth in inches. Use the formula: \(\text{Total Growth} = \text{Current Height} - \text{Height at Birth}.\) Thus, \(66 - 18 = 48\) inches of total growth.

Step 3: Calculate average growth per year in inches

Divide the total growth in inches by the number of years to find the average growth per year. Use the formula: \(\text{Average Growth per Year} = \frac{\text{Total Growth}}{\text{Age}}.\) Hence, \(\frac{48}{20} = 2.4\) inches per year.

Step 4: Convert average growth to centimeters

Convert the average growth per year from inches to centimeters using the conversion factor \(1\) inch equals \(2.54\) cm. Calculate: \(2.4 \times 2.54 = 6.096\) cm per year.

Key Concepts

Unit Conversion

Unit conversion is an essential skill in many scientific and everyday calculations. It allows you to work easily across different measurement systems. In our problem, we start by converting height measurements from a mix of feet and inches into inches alone. This simplifies calculations and helps us compare like with like.
To convert feet to inches, remember that one foot contains 12 inches. Multiply the number of feet by 12 and add the remaining inches. For example, a height of 5 feet 6 inches converts to 66 inches. This ensures all your measurements are in the same units.
Next, we need to convert measurements from inches to centimeters. For this, use the conversion factor: 1 inch is equivalent to 2.54 cm. Multiply the number of inches by 2.54 to find the length in centimeters, as seen when we found that 2.4 inches converts to roughly 6.096 cm. Understanding these conversions will aid in solving problems that require switching between metric and imperial systems.

Linear Growth Rate

The linear growth rate is the steady change in height, length, or another measurement over time. It assumes that this change is even and consistent year to year. Such calculations are straightforward if the parameters remain constant.
In our exercise, the average growth rate is found by subtracting the starting measurement (birth height) from the ending measurement (current height). This total growth over a given period allows us to determine how much growth occurred on average each year.
To find the average growth rate, use the formula:

• Total Growth Rate = Ending Height - Starting Height
• Average Growth Per Year = Total Growth Rate / Number of Years

In this specific problem, Natalie’s height has grown by 48 inches over 20 years, which we calculate to an average growth of 2.4 inches per year. Knowing how to calculate the linear growth rate can apply to multiple fields like biology, finance, and more.

Problem Solving

Problem-solving is a vital skill that aids in understanding and tackling problems effectively by breaking them into small, manageable steps. In this problem, we approached the situation methodically:

• Firstly, calculate the current height in a single unit (inches).
• Secondly, determine the total change in height.
• Next, find the average growth rate over a period.
• Finally, convert the calculated average growth into different measurement units.

By tackling each part individually, it's easier to check your work and avoid errors. This problem-solving technique allows you to systematically solve each step ensuring a clear path to the solution. Each calculation or conversion should be checked to ensure accuracy. Whether you are dealing with a math problem or a real-world issue, structured problem-solving helps you reach the goal comprehensively.