A jogger runs at an average speed of \(5.9 \mathrm{mi} / \mathrm{h}\). (a) How fast is she running in \(\mathrm{m} / \mathrm{s} ?\) (b) How many kilometers does she run in \(98 \min ?(c)\) If she starts a run at \(11: 15 \mathrm{am},\) what time is it after she covers \(4.75 \times 10^{4} \mathrm{ft} ?\)

Short Answer

Expert verified
(a) \(2.639 \text{ m/s}\), (b) \(15.505 \text{ km}\), (c) \(12:46 \text{ PM}\)

Step by step solution

01

Convert Speed to Meters per Second

First, convert the speed from miles per hour to meters per second. Use the conversion factors: 1 mile = 1609.34 meters and 1 hour = 3600 seconds. Convert 5.9 miles per hour to meters per second:\[ 5.9 \text{ mi/h} \times \frac{1609.34 \text{ m}}{1 \text{ mi}} \times \frac{1 \text{ h}}{3600 \text{ s}} \approx 2.639 \text{ m/s} \]
02

Calculate Distance in Kilometers

Calculate the distance the jogger can cover in 98 minutes. First, convert 98 minutes to hours:\[ 98 \text{ min} \times \frac{1 \text{ h}}{60 \text{ min}} = 1.6333 \text{ h} \]Then, use the speed in miles per hour to find the distance in miles, and convert to kilometers (1 mile = 1.60934 kilometers):\[ 5.9 \text{ mi/h} \times 1.6333 \text{ h} = 9.63647 \text{ mi} \]Convert to kilometers:\[ 9.63647 \text{ mi} \times 1.60934 \text{ km/mi} \approx 15.505 \text{ km} \]
03

Time to Cover Given Distance in Feet

First, convert the distance from feet to miles using the conversion factor 1 mile = 5280 feet:\[ 4.75 \times 10^{4} \text{ ft} \times \frac{1 \text{ mi}}{5280 \text{ ft}} \approx 8.992 \text{ mi} \]Then, determine the time it takes to run this distance at the speed of 5.9 miles per hour:\[ \frac{8.992 \text{ mi}}{5.9 \text{ mi/h}} \approx 1.524 \text{ h} \]Convert 1.524 hours to minutes:\[ 1.524 \text{ h} \times 60 \text{ min/h} \approx 91.44 \text{ min} \]Add this time to the start time of 11:15 AM to find the end time:\[ 11:15 \text{ AM} + 91.44 \text{ min} = 12:46 \text{ PM} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Meter per Second Conversion
When converting units, it’s crucial to use the correct conversion factors. For distance, we know that 1 mile equals 1609.34 meters. For time, 1 hour equals 3600 seconds. These conversion factors help us convert joggers' speed from miles per hour \(\text{mi/h}\) to meters per second \(\text{m/s}\). For instance, if a jogger runs at 5.9 \(\text{mi/h}\), converting to \(\text{m/s}\) looks like this:
\[ 5.9 \times \frac{1609.34}{3600} \approx 2.639 \ \text{m/s} \] This conversion helps standardize speed measurements, making calculations straightforward across different units. It’s important to always double-check that you’re using the correct conversion factors to ensure accurate results.
Distance Calculation
Calculating distance involves understanding both speed and time. Begin by converting given time units if necessary. For example, 98 minutes converts to hours as follows:
\[ 98 \times \frac{1}{60} \approx 1.6333 \ \text{h} \] Next, use the speed \(\text{mi/h}\) to find the distance in miles:
\[ 5.9 \times 1.6333 \approx 9.63647 \ \text{mi} \] Finally, convert this distance to kilometers using the conversion factor (1 mile = 1.60934 kilometers):
\[ 9.63647 \times 1.60934 \approx 15.505 \ \text{km} \] This systematic approach ensures we precisely calculate the distance covered in a specified time period.
Time Calculation
Calculating the time to cover a certain distance requires understanding the relationship between distance, speed, and time. First, if distance is provided in feet, convert it to miles (1 mile = 5280 feet):
\[ \frac{4.75 \times 10^4}{5280} \approx 8.992 \ \text{mi} \] Next, use the jogger's speed \(\text{mi/h}\) to find the time in hours:
\[ \frac{8.992}{5.9} \approx 1.524 \ \text{h} \] Convert this time from hours to minutes:
\[ 1.524 \times 60 \approx 91.44 \ \text{min} \] Finally, add 91.44 minutes to the start time of 11:15 AM to determine the end time, converting minutes to hours and minutes as needed. This results in approximately 12:46 PM. Always ensure to convert between units properly to get accurate time calculations.

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Most popular questions from this chapter

Liquid nitrogen is obtained from liquefied air and is used industrially to prepare frozen foods. It boils at \(77.36 \mathrm{~K}\). (a) What is this temperature in \({ }^{\circ} \mathrm{C} ?\) (b) What is this temperature in \({ }^{\circ} \mathrm{F}\) ? (c) At the boiling point, the density of the liquid is \(809 \mathrm{~g} / \mathrm{L}\) and that of the gas is \(4.566 \mathrm{~g} / \mathrm{L}\). How many liters of liquid nitrogen are produced when \(895.0 \mathrm{~L}\) of nitrogen gas is liquefied at \(77.36 \mathrm{~K} ?\)

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