Problem 62 - Algebra A Combined Function

Question

On October 4,2004, the rocket plane SpaceShipOne shot to an altitude of more than \(100 \mathrm{~km}\) for the second time inside a week to claim the \(\$ 10\) million Ansari \(\mathrm{X}\) -Prize. At one point in its flight, SpaceShipOne was traveling past Mach \(1,\) about 930 miles per hour. Find this speed in feet per second. (Source: Space.com)

Step-by-Step Solution

Answer

1363.44 feet per second.

Step 1: Understand the Given Values

We are given that SpaceShipOne was traveling at a speed of 930 miles per hour. Our task is to convert this speed into feet per second.

Step 2: Convert Miles to Feet

We know that 1 mile is equal to 5280 feet. Therefore, to convert 930 miles to feet, we multiply: \( 930 \text{ miles} \times 5280 \text{ feet/mile} \).

Step 3: Convert Hours to Seconds

1 hour is equal to 3600 seconds. Thus, we need to convert the speed from feet per hour to feet per second. We will divide the result from Step 2 by 3600 seconds/hours.

Step 4: Calculate the Speed in Feet per Second

First, execute the multiplication to convert miles to feet: \( 930 \times 5280 = 4,910,400 \text{ feet} \). Then, divide by the number of seconds in an hour to get the speed in feet per second: \( \frac{4,910,400 \text{ feet}}{3600 \text{ seconds}} \approx 1363.44 \text{ feet/second} \).

Step 5: Present the Result

SpaceShipOne's speed of 930 miles per hour is approximately 1363.44 feet per second.

Key Concepts

Miles per Hour to Feet per Second

Understanding unit conversion is key to solve any problem involving different measurement systems. When we need to compare or interpret speeds in different units, converting between miles per hour (mph) and feet per second (ft/s) is common.

• Miles per hour (mph) is a measurement of how many miles a moving object covers in one hour.
• Feet per second (ft/s) calculates how many feet a moving object travels in a second.

To convert mph to ft/s, follow these steps: 1. **Convert miles to feet**: Since 1 mile equals 5280 feet, multiply the speed in miles per hour by 5280. 2. **Convert hours to seconds**: As there are 3600 seconds in an hour, divide the result by 3600.
In our given example, SpaceShipOne's speed was 930 mph. Therefore, we first multiply 930 by 5280 to find the speed in feet per hour. The result is then divided by 3600 to convert into feet per second. This gives the final speed of approximately 1363.44 ft/s.

Speed Calculation

Speed calculation is a fundamental concept in physics and mathematics, utilized to determine how fast an object is moving. When converting units or calculating speeds, keeping track of your given formulas and conversions is crucial.

• The basic speed formula is given by: \[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \]

In our situation with SpaceShipOne:- **Distance** refers to the total distance, which when converted becomes 4,910,400 feet.- **Time** is always noted in hours or seconds, which after conversion is 3600 seconds for one hour.
By employing the speed formula and converting the units, we reach our converted speed in feet per second. Hence, 1363.44 ft/s is the calculated speed, using the process of multiplication and division properly.

Algebraic Problem-Solving

Algebraic problem-solving involves breaking down the problem into manageable parts, using known formulas and constants to obtain the solution. In this context, it requires a logical flow of operations and unit conversions. When tackling the problem of converting 930 mph to ft/s, the step-by-step solution involves:

• Breaking down the conversion of miles to feet.
• Converting hours into seconds to maintain time consistency.
• Applying operations such as multiplication and division in sequence.

This kind of problem-solving relies heavily on understanding the relationships between units and ensures each conversion is correct. By following these steps and confirming each part, the algebraic method helps reach precise answers, such as identifying SpaceShipOne's speed as 1363.44 feet per second. Solutions like these highlight the use of algebra in real-world applications, allowing you to handle more complex conversions and calculations effortlessly.