/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 An 18 -year-old runner can compl... [FREE SOLUTION] | 91Ó°ÊÓ

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An 18 -year-old runner can complete a \(10.0-\mathrm{km}\) course with an average speed of \(4.39 \mathrm{m} / \mathrm{s} .\) A 50 -year-old runner can cover the same distance with an average speed of \(4.27 \mathrm{m} / \mathrm{s} .\) How much later (in seconds) should the younger runner start in order to finish the course at the same time as the older runner?

Short Answer

Expert verified
The younger runner should start approximately 63.26 seconds later.

Step by step solution

01

Calculate Time for 18-year-old Runner

To find the time it takes for the 18-year-old runner to complete the 10 km course, we will use the formula: \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \). The distance is 10 km, which equals 10,000 m. The speed is 4.39 m/sdefine:\[\text{Time}_{18} = \frac{10,000}{4.39} \approx 2277.95 \text{ seconds}\]
02

Calculate Time for 50-year-old Runner

Using the same formula for the 50-year-old runner:\[\text{Time}_{50} = \frac{10,000}{4.27} \approx 2341.21 \text{ seconds}\]
03

Determine the Time Difference

The time difference between the two runners finishing is simply the difference in their completion times:\[\text{Time difference} = \text{Time}_{50} - \text{Time}_{18} = 2341.21 - 2277.95 \approx 63.26 \text{ seconds}\]
04

Interpretation of Time Difference

The younger runner should start approximately 63.26 seconds after the older runner to finish at the same time.

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Key Concepts

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

Average Speed Calculation
Calculating average speed is a straightforward but crucial concept in physics, particularly when dealing with movement and time. Average speed is defined as the total distance traveled divided by the total time taken to travel that distance. Mathematically, it is expressed as: \\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \] It's important to remember that speed is scalar, meaning it only accounts for magnitude and not direction. In our example, both the 18-year-old and 50-year-old runners have set speeds through the 10 km race. The younger runner travels with an average speed of 4.39 m/s and the older at 4.27 m/s. By applying the formula to each runner, you can uncover insightful comparisons on performance characteristics. Understanding how to accurately calculate average speed allows for precise performance tracking over defined distances.
Distance-Time Relationship
The distance-time relationship is foundational to understanding motion. It helps us describe how distance and time interrelate as an object moves. Given this relationship, we typically use the formula for calculating time: \\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \] In this scenario, the runners tackle the same distance of 10 km, equal to 10,000 meters. The time each runner takes can be determined by dividing the distance by their respective speeds. For instance, the 18-year-old's time is calculated as approximately 2277.95 seconds, while the 50-year-old spends about 2341.21 seconds.
  • These calculations highlight the concept that when speed decreases, the time needed to cover a constant distance increases.
  • Such a relationship forms the basis of many physics problems and is key in planning time-sensitive tasks involving travel.
Age-Related Performance Comparison
Age can significantly influence physical performance, as demonstrated by the runners' different speeds. It's important to consider physiological changes that naturally occur over time. Physical capabilities like endurance and strength may decline with age. This explains the slightly slower pace of the 50-year-old runner compared to the 18-year-old runner. Interestingly, calculating the difference in their completion times—approximately 63.26 seconds—provides useful data for assessing age-specific performance. Such analyses offer meaningful insight into how training and age-related factors interplay in athletic settings.
  • Recognizing these differences assists in setting realistic expectations for different age groups.
  • Age-appropriate training and rest regimes are crucial to optimize performance across the lifespan.
Understanding these performance variables helps individuals better grasp how their abilities might change over time and adapt their activities accordingly.

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Most popular questions from this chapter

A VW Beetle goes from 0 to \(60.0 \mathrm{mi} / \mathrm{h}\) with an acceleration of \(+2.35 \mathrm{m} / \mathrm{s}^{2}\) (a) How much time does it take for the Beetle to reach this speed? (b) A top- fuel dragster can go from 0 to \(60.0 \mathrm{mi} / \mathrm{h}\) in \(0.600 \mathrm{s}\). Find the acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ) of the dragster.

A jogger accelerates from rest to \(3.0 \mathrm{m} / \mathrm{s}\) in \(2.0 \mathrm{s}\). A car accelerates from 38.0 to \(41.0 \mathrm{m} / \mathrm{s}\) also in \(2.0 \mathrm{s}\). (a) Find the acceleration (magnitude only) of the jogger. (b) Determine the acceleration (magnitude only) of the car. (c) Does the car travel farther than the jogger during the 2.0 s? If so, how much farther?

The greatest height reported for a jump into an airbag is \(99.4 \mathrm{m}\) by stuntman Dan Koko. In 1948 he jumped from rest from the top of the Vegas World Hotel and Casino. He struck the airbag at a speed of \(39 \mathrm{m} / \mathrm{s}\) \((88 \mathrm{mi} / \mathrm{h}) .\) To assess the effects of air resistance, determine how fast he would have been traveling on impact had air resistance been absent.

The space shuttle travels at a speed of about \(7.6 \times 10^{3} \mathrm{m} / \mathrm{s} .\) The blink of an astronaut's eye lasts about \(110 \mathrm{ms}\). How many football fields (length \(=91.4 \mathrm{m})\) does the shuttle cover in the blink of an eye?

A dynamite blast at a quarry launches a chunk of rock straight upward, and 2.0 s later it is rising at a speed of \(15 \mathrm{m} / \mathrm{s}\). Assuming air resistance has no effect on the rock, calculate its speed (a) at launch and (b) \(5.0 \mathrm{s}\) after launch.

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