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Chapter 2: Problem 70

A runner is at the position \(x=0 \mathrm{m}\) when time \(t=0 \mathrm{s}\). One hundred meters away is the finish line. Every ten seconds, this runner runs half the remaining distance to the finish line. During each ten-second segment, the runner has a constant velocity. For the first forty seconds of the motion, construct (a) the position-time graph and (b) the velocity-time graph.

Short Answer

Expert verified
For the first 40 seconds, the position goes from 0 to 93.75m and velocity from 5 to 0.625m/s. Position decreases exponentially; velocity halves each interval.

Step by step solution

01

Describe Initial Conditions

The runner starts at the position \( x = 0 \text{ m} \) at time \( t = 0 \text{ s} \). The finish line is at \( x = 100 \text{ m} \). In each interval of 10 seconds, the runner covers half of the remaining distance to the finish line, moving with a constant velocity within each interval.
02

Determine Position at Each 10 Second Interval

For the first interval (0-10 seconds), the runner covers half of the initial 100m, ending at 50m. For the second interval (10-20 seconds), the runner covers half of the remaining 50m, ending at 75m. For the third interval (20-30 seconds), the runner covers half of the remaining 25m, ending at 87.5m. For the fourth interval (30-40 seconds), the runner covers half of the remaining 12.5m, ending at 93.75m.
03

Construct the Position-Time Graph

Using the positions calculated, plot the time (seconds) on the x-axis and the position (meters) on the y-axis:- At \( t = 0 \), \( x = 0 \text{ m} \)- At \( t = 10 \), \( x = 50 \text{ m} \)- At \( t = 20 \), \( x = 75 \text{ m} \)- At \( t = 30 \), \( x = 87.5 \text{ m} \)- At \( t = 40 \), \( x = 93.75 \text{ m} \)Connect the points linearly as the velocity within each interval is constant.
04

Calculate Velocity for Each Interval

The velocity is constant for each segment and is calculated as the change in position divided by the time interval (10 seconds):- From 0 to 10 seconds: \( v = \frac{50 \text{ m}}{10 \text{ s}} = 5 \text{ m/s} \)- From 10 to 20 seconds: \( v = \frac{25 \text{ m}}{10 \text{ s}} = 2.5 \text{ m/s} \)- From 20 to 30 seconds: \( v = \frac{12.5 \text{ m}}{10 \text{ s}} = 1.25 \text{ m/s} \)- From 30 to 40 seconds: \( v = \frac{6.25 \text{ m}}{10 \text{ s}} = 0.625 \text{ m/s} \)
05

Construct the Velocity-Time Graph

Using the calculated velocities, plot time (seconds) on the x-axis and velocity (m/s) on the y-axis:- From \( t = 0 \text{ to } 10 \text{ s}, v = 5 \text{ m/s} \)- From \( t = 10 \text{ to } 20 \text{ s}, v = 2.5 \text{ m/s} \)- From \( t = 20 \text{ to } 30 \text{ s}, v = 1.25 \text{ m/s} \)- From \( t = 30 \text{ to } 40 \text{ s}, v = 0.625 \text{ m/s} \)Draw horizontal lines for each interval as the velocity is constant.

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Key Concepts

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

Position-Time Graph
A position-time graph visually represents how a runner's position changes over time. It is a useful tool in kinematics to understand motion. In our example, the runner starts at the origin with position at 0 meters and moves towards the finish line located at 100 meters.
Each point on the graph shows the runner鈥檚 position at a specific time:
  • Starting at time 0 seconds at position 0 meters.
  • At 10 seconds, reaching 50 meters.
  • At 20 seconds, arriving at 75 meters.
  • At 30 seconds, at 87.5 meters.
  • Finally at 40 seconds, reaching 93.75 meters.
Connect these points with straight lines to indicate constant velocity in each segment. The graph shows us that the runner is covering less distance with each subsequent interval.
Velocity-Time Graph
A velocity-time graph illustrates the runner鈥檚 speed at different periods during their journey. Each horizontal line on the graph represents constant velocity over time.
For this runner:
  • From 0 to 10 seconds, the velocity is 5 m/s.
  • From 10 to 20 seconds, the velocity decreases to 2.5 m/s.
  • From 20 to 30 seconds, it further decreases to 1.25 m/s.
  • From 30 to 40 seconds, it is 0.625 m/s.
These velocities reflect how the runner covers half of the remaining distance each time, hence the speed reduces as less distance is left to cover. The horizontal lines indicate that the velocity within each segment is constant.
Constant Velocity
Constant velocity implies that both the speed and direction remain unchanged over a specific time period. In each 10-second segment of the runner鈥檚 journey, the velocity is constant.
This is why we have straight lines on both the position-time and velocity-time graphs during each interval. The constant velocity simplifies calculations: once the velocity is established over an interval, it remains unchanged until the interval expires.
  • In the first segment, the runner moves with a steady pace of 5 m/s.
  • This steady pace decreases as each successive interval has less absolute change in position.
Constant velocity enables us to predict the runner's position at any given time within each segment.
Distance-Time Relationship
The distance-time relationship tells us how far an object travels within a given time frame. For our runner, this relationship is crucial in determining the position at the end of each segment.
The formula to consider is:
  • Distance = Velocity 脳 Time
In the exercise, by knowing the distance covered in each time frame, we can easily plot the position-time graph points. Over the entire 40 seconds, each segment halves the remaining distance:
  • First 10 seconds: 50 meters
  • Next 10 seconds: 25 meters
  • Then 12.5 meters
  • And finally 6.25 meters
Understanding this relationship allows us to see how the runner鈥檚 position approaches the finish line exponentially closer over time.

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Most popular questions from this chapter

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 \(2.0 \mathrm{s} ?\) If so, how much farther?

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(a) Suppose that a NASCAR race car is moving to the right with a constant velocity of \(+82 \mathrm{m} / \mathrm{s}\). What is the average acceleration of the car? (b) Twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed. What is the average acceleration of the car?

Two motorcycles are traveling due east with different velocities. However, four seconds later, they have the same velocity. During this four-second interval, cycle A has an average acceleration of \(2.0 \mathrm{m} / \mathrm{s}^{2}\) due east, while cycle \(\overrightarrow{\mathrm{B}}\) has an average acceleration of \(4.0 \mathrm{m} / \mathrm{s}^{2}\) due east. By how much did the speeds differ at the beginning of the four-second interval, and which motorcycle was moving faster?

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