Math Problem Statement

Jan stands 5 m away from a motion sensor and then walks 4 m toward it at a constant rate for 5 s. Then she walks 2 m away from the location where she changed direction at a variable rate for the next 3 s. She stops and waits at this location for 2 s. Draw a distance versus time graph to show Jan’s motion sensor walk.

Solution

Let's outline Jan's motion based on the description:

  1. Jan starts at 5 meters from a motion sensor and walks 4 meters toward it over a period of 5 seconds.
  2. She then walks 2 meters away from the motion sensor over the next 3 seconds.
  3. Jan stops and waits for 2 seconds.

Here's how to draw the distance versus time graph for Jan's motion:

  • Start at 5 meters at t=0t = 0 seconds.
  • At t=5t = 5 seconds, Jan walks 4 meters towards the sensor, reaching 54=15 - 4 = 1 meter from the sensor.
  • From t=5t = 5 to t=8t = 8 seconds, she walks 2 meters away from the sensor, moving to 1+2=31 + 2 = 3 meters from the sensor.
  • From t=8t = 8 to t=10t = 10 seconds, Jan remains stationary at 3 meters from the sensor.

The graph should look like a piecewise linear plot:

  • Decrease from 5 meters to 1 meter in the first 5 seconds (slope: -0.8 meters/second).
  • Increase from 1 meter to 3 meters between 5 and 8 seconds (slope: +0.67 meters/second).
  • Remain flat at 3 meters from 8 to 10 seconds.

Would you like to see a visual representation of this graph? Here are some further questions and tips related to the concept:

  1. What does the slope of each segment on a distance-time graph represent?
  2. How would the graph change if Jan walked faster or slower during each segment?
  3. What is the significance of the slope changing signs on a distance-time graph?
  4. How can we determine Jan's average speed during her walk?
  5. What would the graph look like if Jan decided to walk in a continuous loop rather than stopping?

Tip: When analyzing motion on a distance-time graph, note that a steeper slope indicates faster motion, while a flat line indicates no motion.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Distance-Time Graph
Rates of Change

Formulas

Distance = Speed × Time
Slope = Δy/Δx (Rate of Change)

Theorems

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Suitable Grade Level

Grade 8-10