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Two people walking on a sidewalk have the following equations of motion: $$ \begin{aligned} &x_{1}=8.2 \mathrm{~m}+(-1.1 \mathrm{~m} / \mathrm{s}) t \\ &x_{2}=5.9 \mathrm{~m}+(1.7 \mathrm{~m} / \mathrm{s}) t \end{aligned} $$ (a) Which person is moving faster? (b) Which person will be at \(x=0\) at some time in the future?

Short Answer

Expert verified
Person 2 is faster. Only Person 1 will be at \(x=0\) in the future.

Step by step solution

01

Understanding Velocity

The velocity of each person can be determined by looking at the coefficient of time \(t\) in their equations. \(x_{1}=8.2 \, m+(-1.1 \, m/s) t\) indicates Person 1 has a velocity of \(-1.1 \, m/s\).\(x_{2}=5.9 \, m+(1.7 \, m/s) t\) indicates Person 2 has a velocity of \(1.7 \, m/s\).
02

Comparing Velocities

To find out who is faster, compare the absolute values of the velocities: \| -1.1 \, m/s \| = 1.1 \, m/s and \| 1.7 \, m/s \| = 1.7 \, m/s. Since \(1.7 \, m/s\) is greater than \(1.1 \, m/s\), Person 2 is faster.
03

Solving for Zero Position of Person 1

Set the equation for Person 1 equal to zero to find when they will be at \x=0\.\[8.2 \, m + (-1.1 \, m/s) t = 0 \-1.1 \, m/s \, t = -8.2 \, m \t = \frac{-8.2 \, m}{-1.1 \, m/s} \t \approx 7.45 \, s \\]This means Person 1 will be at \x=0\ at \t \approx 7.45 \, s\.
04

Checking for Zero Position of Person 2

Set the equation for Person 2 equal to zero to find when they will be at \x=0\.\[5.9 \, m + (1.7 \, m/s) t = 0 \1.7 \, m/s \, t = -5.9 \, m \t = \frac{-5.9 \, m}{1.7 \, m/s} \t = -3.47 \, s \\]Since time cannot be negative, Person 2 will not be at \x=0\ at any positive time.

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Key Concepts

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

Velocity Comparison
Understanding how to compare velocities is an essential part of solving motion problems. In the context of this exercise, velocity defines how fast a person moves along a path. Based on their equations of motion, person 1 has a velocity of \(-1.1 \, \text{m/s}\) and person 2 has \(1.7 \, \text{m/s}\). Here, the magnitude of the velocity indicates speed, regardless of direction (positive or negative indicates forward or backward motion, respectively).
  • Magnitude of Person 1's velocity: \(|-1.1| = 1.1 \, \text{m/s}\)
  • Magnitude of Person 2's velocity: \(|1.7| = 1.7 \, \text{m/s}\)
Since \(1.7 \, \text{m/s}\) is greater than \(1.1 \, \text{m/s}\), we conclude that Person 2 is indeed moving faster. Always focus on the magnitude when comparing velocities, as direction is often secondary in speed determination.
Position Calculation
Calculating the position of an object or person involves using the equation of motion. Position is an important factor because it helps determine where each person is at any given time. For Person 1, the equation for position is given by \(x_{1}=8.2 \, \text{m}+(-1.1 \, \text{m/s}) t\). To find when Person 1 will be at \(x=0\), we set the equation to zero:\[8.2 \, \text{m} + (-1.1 \, \text{m/s}) t = 0\]Solving for \(t\),\[-1.1 \, \text{m/s} \, t = -8.2 \, \text{m}\] \[t = \frac{-8.2 \, \text{m}}{-1.1 \, \text{m/s}} = 7.45 \, \text{s}\]Person 1 reaches \(x=0\) after approximately \(7.45\) seconds. For Person 2's position equation, \(x_{2}=5.9 \, \text{m}+(1.7 \, \text{m/s}) t\), the same approach shows they will not cross \(x=0\) at a meaningful (positive) time, given the negative result, which is outside our realistic timeline.
Linear Motion Analysis
Linear motion analysis is the study of objects moving along a straight path. It provides tools to analyze how position and velocity change over time. In this case, Person 1 and Person 2 have straight-line paths with specific starting points and velocities.Key aspects include:
  • Direction: Both move in opposite directions; Person 1 moves backward while Person 2 moves forward.
  • Velocity: As derived from their equations, velocities determine how fast they change their respective positions over time.
  • Time: Evaluating at specific moments, like \(t=0\) or when \(x=0\), reveals insights about their motion scenarios.
Considering linear motion, analyzing these aspects helps determine future positions and compare how quickly each person gets to a new location or reference point. This clear understanding of their velocities and resulting positions can be crucial for solving real-world problems where timing and path tracking are vital.

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Most popular questions from this chapter

Think \& Calculate A train travels in a straight line at \(20.0 \mathrm{~m} / \mathrm{s}\) for \(2 \mathrm{~km}\), then at \(30.0 \mathrm{~m} / \mathrm{s}\) for another \(2 \mathrm{~km}\). (a) Is the average speed of the train greater than, less than, or equal to \(25 \mathrm{~m} / \mathrm{s}\) ? Explain. (b) Verify your answer to part (a) by calculating the average speed.

Cleo the black lab runs to pick up a stick on the ground at the location \(x=3.0 \mathrm{~m}\). The equation of motion for Cleo is \(x=-12.1 \mathrm{~m}+(5.2 \mathrm{~m} / \mathrm{s})\) t. (a) Where is Cleo at \(t=1.6 \mathrm{~s}\) ? (b) At what time does Cleo reach the stick?

A rose-covered parade float is at \(x=0\) at time \(t=0\). The float moves in a straight line at \(2.0 \mathrm{~m} / \mathrm{s}\) for the next \(5 \mathrm{~s}\) before coming to a stop. After a 5 -s stop, the float moves again at \(1.0 \mathrm{~m} / \mathrm{s}\) in the same direction as before. (a) Sketch the position-time graph for the float from the time \(t=0\) until the time \(t=15 \mathrm{~s}\). (b) From your graph, determine the positions of the float at \(t=2 \mathrm{~s}\) and \(t=11 \mathrm{~s}\).

Triple Choice You travel along the \(x\) axis from the location \(x_{\mathrm{i}}=10 \mathrm{~m}\) to the location \(x_{\mathrm{f}}=25 \mathrm{~m}\). Your location \(x_{\mathrm{i}}=10 \mathrm{~m}\) to the location \(x_{\mathrm{f}}=25 \mathrm{~m}\). Your friend travels from \(x_{\mathrm{i}}=35 \mathrm{~m}\) to \(x_{\mathrm{f}}=40 \mathrm{~m}\). Is the distance you cover greater than, less than, or equal to the distance covered by your friend? Explain.

The velocity of an object that moves with constant velocity is increased. Does this change the intercept or the slope of the position-time graph of the object's motion? Explain.

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