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Chapter 2: Problem 4

An 18 -year-old runner can complete a \(10.0\) -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 65 seconds later to finish at the same time.

Step by step solution

01

Calculate Completion Time for Younger Runner

The younger runner's speed is given as \(4.39\, \text{m/s}\). The total distance of the course is \(10.0\, \text{km} = 10,000\, \text{m}\). To find the time it takes, use the formula \(\text{time} = \frac{\text{distance}}{\text{speed}}\). Thus, the time taken by the younger runner is: \[\text{time} = \frac{10000}{4.39} \approx 2277.45\, \text{seconds}. \]
02

Calculate Completion Time for Older Runner

The older runner's speed is given as \(4.27\, \text{m/s}\). We use the same formula to calculate the time: \(\text{time} = \frac{\text{distance}}{\text{speed}}\). Thus, the time taken by the older runner is: \[\text{time} = \frac{10000}{4.27} \approx 2342.83\, \text{seconds}. \]
03

Determine the Time Difference

Now, find the difference in their completion times to determine how much later the younger runner should start. Subtract the younger runner's time from the older runner's time: \[\text{time difference} = 2342.83 - 2277.45 = 65.38\, \text{seconds}.\]
04

Round Time Difference

Since runners cannot start in fractions of a second, round the time difference to the nearest whole number, which gives \(65\) seconds.

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Key Concepts

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

Distance Calculation
When dealing with motion problems, understanding and working with distance is a fundamental skill. Distance is typically measured in meters or kilometers, depending on the scale of the problem.
In this scenario, both runners are covering a distance of 10 kilometers, which we convert into meters for consistency in calculations:
  • 1 kilometer = 1,000 meters
So, 10 kilometers becomes 10,000 meters.
This conversion is essential to ensure the units of distance and speed align, allowing accurate computation of time, speed, or any other related measurement.
Average Speed
Average speed is a key element in motion problems, describing how fast an object travels over a specific distance.
It is calculated by dividing the total distance by the total time taken to cover it:
  • Formula: \( \text{average speed} = \frac{\text{total distance}}{\text{total time}} \)
In our exercise, the average speed is given for both runners. For the younger runner, it's \(4.39\, \text{m/s}\), and for the older runner, it's \(4.27\, \text{m/s}\). Knowing these speeds helps us determine how much time each takes to complete the course.
Time Difference
In motion problems, understanding the time difference between events is crucial. It's the difference in the time taken by two entities to complete the same task.
To derive the time difference, calculate the time each runner takes to cover the same distance using the formula:
  • \( \text{time} = \frac{\text{distance}}{\text{speed}} \)
For the younger runner, the time is approximately 2277.45 seconds, while for the older, it's about 2342.83 seconds. The time difference is:
  • \( 2342.83 - 2277.45 = 65.38 \) seconds
Rounding this gives 65 seconds, indicating how much later the younger runner should start to finish together with the older runner.
Motion Problems
Motion problems, like the one in our exercise, frequently ask us to compute speeds, times, or distances to understand how entities move relative to each other.
These problems are a practical application of all the concepts discussed:
  • Distance Calculation: Converting units like kilometers to meters.
  • Average Speed: Figuring out how fast something is moving over time.
  • Time Difference: Calculating how timing will affect different objects moving towards a goal.
In such scenarios, remember to align all your units, carefully employ formulas, and account for practical elements such as rounding, as seen in rounding the time difference. This ensures that your solution is both accurate and realistic.

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Most popular questions from this chapter

A penny is dropped from rest from the top of the Sears Tower in Chicago. Considering that the height of the building is \(427 \mathrm{~m}\) and ignoring air resistance, find the speed with which the penny strikes the ground.

An automobile starts from rest and accelerates to a final velocity in two stages along a straight road. Each stage occupies the same amount of time. In stage 1 , the magnitude of the car's acceleration is \(3.0 \mathrm{~m} / \mathrm{s}^{2}\). The magnitude of the car's velocity at the end of stage 2 is \(2.5\) times greater than it is at the end of stage 1 . Find the magnitude of the acceleration in stage 2 .

A speed ramp at an airport is basically a large conveyor belt on which you can stand and be moved along. The belt of one ramp moves at a constant speed such that a person who stands still on it leaves the ramp 64 s after getting on. Clifford is in a real hurry, however, and skips the speed ramp. Starting from rest with an acceleration of \(0.37\) \(\mathrm{m} / \mathrm{s}^{2}\), he covers the same distance as the ramp does, but in one-fourth the time. What is the speed at which the belt of the ramp is moving?

A whale swims due east for a distance of \(6.9 \mathrm{~km},\) turns around and goes due west for \(1.8 \mathrm{~km},\) and finally turns around again and heads \(3.7 \mathrm{~km}\) due east. (a) What is the total distance traveled by the whale? (b) What are the magnitude and direction of the displacement of the whale?

Interactive Solution \(2.31\) at offers help in modeling this problem. A car is traveling at a constant speed of \(33 \mathrm{~m} / \mathrm{s}\) on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is \(2.5 \mathrm{~km}\) away?
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