Math Problem Statement

Calculate the average velocity of the object in motion in the following time intervals: 0-2 s, 2-3 s, 3-5 s, and 5-8 s, as well as the overall average velocity.

Solution

To calculate the average velocity for each interval, we use the formula for average velocity:

vavg=ΔxΔtv_{\text{avg}} = \frac{\Delta x}{\Delta t}

where:

  • Δx\Delta x is the change in position (displacement),
  • Δt\Delta t is the change in time.

Let's analyze each time interval using the graph you've provided:

1. Between 0 and 2 seconds:

From the graph:

  • At t=0t = 0, the position is 00 meters.
  • At t=2t = 2, the position is 88 meters.

So: vavg=8m0m2s0s=82=4m/sv_{\text{avg}} = \frac{8 \, \text{m} - 0 \, \text{m}}{2 \, \text{s} - 0 \, \text{s}} = \frac{8}{2} = 4 \, \text{m/s}

2. Between 2 and 3 seconds:

From the graph:

  • At t=2t = 2, the position is 88 meters.
  • At t=3t = 3, the position is still 88 meters (no change in position).

So: vavg=8m8m3s2s=01=0m/sv_{\text{avg}} = \frac{8 \, \text{m} - 8 \, \text{m}}{3 \, \text{s} - 2 \, \text{s}} = \frac{0}{1} = 0 \, \text{m/s}

3. Between 3 and 5 seconds:

From the graph:

  • At t=3t = 3, the position is 88 meters.
  • At t=5t = 5, the position is 2020 meters.

So: vavg=20m8m5s3s=122=6m/sv_{\text{avg}} = \frac{20 \, \text{m} - 8 \, \text{m}}{5 \, \text{s} - 3 \, \text{s}} = \frac{12}{2} = 6 \, \text{m/s}

4. Between 5 and 8 seconds:

From the graph:

  • At t=5t = 5, the position is 2020 meters.
  • At t=8t = 8, the position is 2828 meters.

So: vavg=28m20m8s5s=832.7m/sv_{\text{avg}} = \frac{28 \, \text{m} - 20 \, \text{m}}{8 \, \text{s} - 5 \, \text{s}} = \frac{8}{3} \approx 2.7 \, \text{m/s}

Part b: Average velocity for the overall motion (from 0 to 8 seconds):

From the graph:

  • The total displacement from t=0t = 0 to t=8t = 8 is:
    • At t=0t = 0, position is 00 meters.
    • At t=8t = 8, position is 2828 meters.

So the total displacement is 28m28 \, \text{m} and the total time is 8s8 \, \text{s}.

vavg=28m8s=3.5m/sv_{\text{avg}} = \frac{28 \, \text{m}}{8 \, \text{s}} = 3.5 \, \text{m/s}


Summary:

  • Between 0 and 2 s: 4m/s4 \, \text{m/s}
  • Between 2 and 3 s: 0m/s0 \, \text{m/s}
  • Between 3 and 5 s: 6m/s6 \, \text{m/s}
  • Between 5 and 8 s: 2.7m/s2.7 \, \text{m/s}
  • Overall average velocity: 3.5m/s3.5 \, \text{m/s}

Do you want further clarification or details on these calculations?


Here are five related questions for deeper understanding:

  1. What would happen if the object had moved backward in any interval?
  2. How does the slope of the graph relate to the velocity in each interval?
  3. What is the difference between instantaneous velocity and average velocity?
  4. How would the graph look if the object moved with constant velocity?
  5. Can velocity be negative? How would that appear on the graph?

Tip: Average velocity depends on the total displacement, not the total distance traveled!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Kinematics
Average Velocity
Position-Time Graph

Formulas

v_avg = Δx / Δt

Theorems

-

Suitable Grade Level

Grades 9-12