Question

In heavy rush-hour traffic you drive in a straight line at \(12 \mathrm{~m} / \mathrm{s}\) for \(1.5 \mathrm{~min}\), then you have to stop for \(3.5 \mathrm{~min}\), and finally you drive at \(15 \mathrm{~m} / \mathrm{s}\) for another \(2.5 \mathrm{~min}\). (a) Plot a position-time graph for this motion. Your graph should extend from \(t=0\) to \(t=7.5 \mathrm{~min}\). (b) Use your graph from part (a) to calculate the average velocity between \(t=0\) and \(t=7.5 \mathrm{~min}\).

Short answer

Plot a position-time graph with linear and horizontal segments. Average velocity is 7.4 m/s between 0 and 7.5 minutes.

Step-by-step solution

Determine Position Over Time Intervals

The movement occurs in three phases: 1. Moving at 12 m/s for 1.5 minutes: - Convert time to seconds: \[ 1.5 \, \text{min} = 90 \, \text{s} \] - Calculate the distance: \[ d_1 = 12 \, \text{m/s} \times 90 \, \text{s} = 1080 \, \text{m} \]2. Stopping for 3.5 minutes, so position remains the same: - Convert time to seconds: \[ 3.5 \, \text{min} = 210 \, \text{s} \] - Position is constant at 1080 m during this interval.3. Moving at 15 m/s for 2.5 minutes: - Convert time to seconds: \[ 2.5 \, \text{min} = 150 \, \text{s} \] - Calculate the distance covered in this phase: \[ d_2 = 15 \, \text{m/s} \times 150 \, \text{s} = 2250 \, \text{m} \]Combine the distances: position after third phase is \[ 1080 \, \text{m} + 2250 \, \text{m} = 3330 \, \text{m} \]

Plot the Position-Time Graph

Using the calculations from Step 1, plot the position over time:- From \(t = 0\) to \(t = 90\) seconds (1.5 min), the position linearly increases from 0 to 1080 m with a slope of 12 m/s.- From \(t = 90\) to \(t = 300\) seconds (5 min), the position remains constant at 1080 m, showing a horizontal line on the graph.- From \(t = 300\) to \(t = 450\) seconds (7.5 min), the position linearly increases from 1080 m to 3330 m with a slope of 15 m/s.This graph should be linear over the moving sections and horizontal during the rest period.

Calculate Average Velocity Using Total Displacement and Total Time

Average velocity is defined as total displacement divided by total time.- Total displacement from \(t = 0\) seconds to \(t = 450\) seconds is 3330 m.- Total time is \[ 1.5 \, \text{min} + 3.5 \, \text{min} + 2.5 \, \text{min} = 7.5 \, \text{min} \] Convert this to seconds: \[ 7.5 \, \text{min} = 450 \, \text{s} \]- Average velocity (\( \bar{v} \)) is: \[ \bar{v} = \frac{\text{Total displacement}}{\text{Total time}} = \frac{3330 \, \text{m}}{450 \, \text{s}} = 7.4 \, \text{m/s} \]

Review the Results

The calculations indicate that the average velocity during the entire journey, from the start to the end, is 7.4 m/s. The position-time graph effectively shows differing motion stages - moving, halting, and then moving again - which helps in visualizing the journey's timeline.

Key concepts

Position-Time Graph

A position-time graph is a useful tool to visually represent the motion of an object over a period of time. It plots position on the y-axis and time on the x-axis. In our exercise, the graph provides a clear picture of the car's journey through periods of driving and stopping.

To create this graph, start by determining the total time spent on each phase of the journey. It's important to convert minutes into seconds to maintain consistency in units as distances are in meters and time is usually measured in seconds in physics.

Here's the breakdown:

• First phase: driving at 12 m/s for 1.5 minutes (or 90 seconds)
• Second phase: stopping for 3.5 minutes (or 210 seconds)
• Third phase: driving again, this time at 15 m/s for 2.5 minutes (or 150 seconds)

With these time segments plotted, you'll see non-linear increases where the car is moving and a flat line during the stop. Such plots help illustrate how different velocities and stoppages affect overall position over time.

Constant Velocity

Constant velocity implies that an object moves at a steady speed in a straight path. In the given exercise, two segments of the journey illustrate constant velocity.During the first phase, the car travels at a steady pace of 12 m/s. You calculate the distance covered by multiplying the velocity by time: \[d = v imes t = 12 ext{ m/s} imes 90 ext{ s} = 1080 ext{ m}\]The motion is uniform here, which means there's no acceleration or deceleration.

Similarly, in the third phase, the car moves at a constant velocity of 15 m/s, resulting in a distance of:\[d = 15 ext{ m/s} imes 150 ext{ s} = 2250 ext{ m}\]The concept of constant velocity is crucial as it simplifies calculations by maintaining a uniform speed without unexpected changes. This aspect brings predictability to the calculations and graph representations.

Displacement

Displacement refers to the change in position of an object and is expressed in meters (m). It's different from distance because displacement considers direction as well.

In the context of this exercise, the total displacement of the car is the difference between its final and initial position after all phases of movement are completed. Here, the car starts from zero and ends up at a position of 3330 meters after the entire journey.

• Phase 1 results in a position of 1080 m.
• In phase 2, the car remains stationary at 1080 m.
• Phase 3 adds another 2250 m, reaching 3330 m.

Each phase contributes to the overall displacement regardless of stops as long as movement occurs in a straight line.

Unit Conversion

Unit conversion is an essential skill in physics, allowing you to express measurements in consistent units. This process is key for accuracy and clarity in calculations.

For this problem, time needed conversion from minutes to seconds to match velocity in meters per second (m/s). Here's how conversions are handled:

• 1.5 minutes to seconds: \[1.5 ext{ min} imes 60 = 90 ext{ seconds}\]
• 3.5 minutes to seconds: \[3.5 ext{ min} imes 60 = 210 ext{ seconds}\]
• 2.5 minutes to seconds: \[2.5 ext{ min} imes 60 = 150 ext{ seconds}\]
• Total journey time of 7.5 minutes: \[7.5 ext{ min} imes 60 = 450 ext{ seconds}\]

Converting units is a simple yet powerful tool that allows for smooth, consistent calculations and helps in preventing errors due to mismatched units.