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Chapter 2: Problem 70

A runner is at the position \(x=0 \mathrm{m}\) when time \(t=0\) 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
The position-time graph is a curve; the velocity-time graph has stepwise decreasing values.

Step by step solution

01

Determine the Position at Each Time Interval

The runner starts at position \(x = 0\) meters and the finish line is at \(x = 100\) meters. Every 10 seconds, the runner covers half the remaining distance to the finish line. Calculate the position at \(t = 10\) s, \(t = 20\) s, \(t = 30\) s, and \(t = 40\) s: - At \(t = 10\) s: The runner covers half of the 100 meters, reaching \(x = 50\) meters. - At \(t = 20\) s: Half of the remaining 50 meters, reaching \(x = 75\) meters. - At \(t = 30\) s: Half of the remaining 25 meters, reaching \(x = 87.5\) meters. - At \(t = 40\) s: Half of the remaining 12.5 meters, reaching \(x = 93.75\) meters.
02

Plot the Position-Time Graph

Using the positions calculated in Step 1, plot the position-time graph. The x-axis represents time in seconds, and the y-axis represents the position in meters. Plot the points: - (0, 0) - (10, 50) - (20, 75) - (30, 87.5) - (40, 93.75) Connect these points to form a curve.
03

Calculate the Velocity for Each Interval

Velocity can be determined by the change in position over time for each interval: - Between \(t = 0\) s and \(t = 10\) s: Velocity is \((50-0)/(10-0) = 5\) m/s. - Between \(t = 10\) s and \(t = 20\) s: Velocity is \((75-50)/(20-10) = 2.5\) m/s. - Between \(t = 20\) s and \(t = 30\) s: Velocity is \((87.5-75)/(30-20) = 1.25\) m/s. - Between \(t = 30\) s and \(t = 40\) s: Velocity is \((93.75-87.5)/(40-30) = 0.625\) m/s.
04

Plot the Velocity-Time Graph

Using the velocities calculated in Step 3, plot the velocity-time graph. The x-axis represents time in seconds, and the y-axis represents velocity in meters per second. Plot the horizontal line segments: - From \(t = 0\) s to \(t = 10\) s at \(v = 5\) m/s. - From \(t = 10\) s to \(t = 20\) s at \(v = 2.5\) m/s. - From \(t = 20\) s to \(t = 30\) s at \(v = 1.25\) m/s. - From \(t = 30\) s to \(t = 40\) s at \(v = 0.625\) m/s.

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Key Concepts

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

velocity-time graph
A velocity-time graph illustrates how an object's velocity changes over a certain time period. The x-axis shows time, while the y-axis displays velocity in meters per second (m/s). For the runner's scenario, the velocity at any given period was constant. This means each 10-second segment can be depicted as a flat line on the velocity-time graph.
  • From 0 to 10 seconds, the runner ran with a velocity of 5 m/s. The graph will have a horizontal line at 5 m/s.
  • Between 10 and 20 seconds, the velocity decreased to 2.5 m/s.
  • From 20 to 30 seconds, it further decreased to 1.25 m/s.
  • Lastly, from 30 to 40 seconds, the velocity was 0.625 m/s.
This gradual decrease in velocity for each subsequent interval creates a descending step pattern on the graph.
Understanding these visuals helps grasp the concept of how velocity changes with time, providing clear insights into the runner's behavior over the intervals.
constant velocity
Constant velocity means that an object is moving in a straight line at a steady speed. During each 10-second interval, the runner maintained a constant velocity, which means the speed didn't change for those specific periods. In terms of physics, constant velocity signifies both uniform speed and direction. In our runner's case, while the direction remained constant (towards the finish line), the actual speed changed between intervals.
  • For each 10-second interval, the runner's velocity was constant, albeit different from the previous interval.
  • Constant velocity implies that on a position-time graph, the slope of the line segment steadily represents the velocity.
Though the velocities were different in successive intervals, within each interval, the runner maintained steady speeds, reaching set positions consistently during each timeframe.
distance-time relationship
The distance-time relationship is crucial to understanding how an object traverses space over time. The position-time graph embodies this relationship, showing how the runner closed the gap to the finish line.
  • Initially, the runner was at 0 meters at time zero.
  • Each subsequent 10-second interval brought the runner closer, covering half the remaining distance each time.
  • For example, from 0-10 seconds, the runner went from 0 to 50 meters.
  • Then, from 10 to 20 seconds, the position reached 75 meters, and so on.
This method of traveling half the remaining distance results in a series of diminishing increments, which when plotted as points on a position-time graph, form a curve that suggests the runner's slowing progress.
Understanding this relationship between time and distance visualizes how the decreasing velocity affects the runner's journey to the finish line.

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Most popular questions from this chapter

A cheetah is hunting. Its prey runs for 3.0 \(\mathrm{s}\) at a constant velocity of \(+9.0 \mathrm{m} / \mathrm{s}\) . Starting from rest, what constant acceleration must the cheetah maintain in order to run the same distance as its prey runs in the same time?

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.0 \(\mathrm{m} / \mathrm{s}\) . The car is a distance \(d\) away. The bear is 26 \(\mathrm{m}\) behind the tourist and running at 6.0 \(\mathrm{m} / \mathrm{s}\) . The tourist reaches the car safely. What is the maximum possible value for \(d ?\)

(a) What is the magnitude of the average acceleration of a skier who, starting from reaches a speed of 8.0 \(\mathrm{m} / \mathrm{s}\) when going down a slope for 5.0 \(\mathrm{s} ? \quad\) (b) How far does the skier travel in this time?

Review Conceptual Example 7 as background for this problem. A car is traveling to the left, which is the negative direction. The direction of travel remains the same throughout this problem. The car's initial speed is \(27.0 \mathrm{m} / \mathrm{s},\) and during a 5.0 \(\mathrm{s}\) - interval, it changes to a final speed of (a) 29.0 \(\mathrm{m} / \mathrm{s}\) and \((\mathrm{b}) 23.0 \mathrm{m} / \mathrm{s}\) . In each case, find the acceleration ( magnitude and algebraic sign) and state whether or not the car is decelerating.

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 \((\mathrm{a})\) at launch and (b) 5.0 \(\mathrm{s}\) after launch.
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