1d Kinematics, missing something.

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SUMMARY

The discussion focuses on calculating average velocity in a 1D kinematics problem involving a position-time graph. The user initially miscalculated the time conversion from hours to seconds, using 300 seconds instead of the correct 720 seconds for 0.2 hours. The formula for average velocity, vave = Δx / Δt, was correctly applied, leading to the resolution of the problem once the time was accurately determined. The user successfully corrected their calculations and confirmed their answers as correct.

PREREQUISITES
  • Understanding of basic kinematics concepts
  • Familiarity with the formula for average velocity (vave = Δx / Δt)
  • Ability to convert time units (hours to seconds)
  • Knowledge of interpreting position-time graphs
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  • Study the principles of 1D kinematics in detail
  • Practice converting between different time units for physics problems
  • Learn how to analyze position-time graphs for various motion scenarios
  • Explore more complex kinematics problems involving acceleration
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This discussion is beneficial for physics students, educators, and anyone interested in mastering the concepts of kinematics and average velocity calculations.

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A person who walks for exercise produces the position-time graph given with this problem.

02_64.gif


(b) Calculate the average velocity for each segment to verify your answers to part (a).
A
m/s
B
m/s
C
m/s
D
m/s

B is obviously 0

For A I did .20 of an hour is 300 seconds.
And 1km = 1000m

So 1000/300 =3.33 and that is wrong..

What am I doing wrong here I used:
[tex]v_{ave} = \Delta x / \Delta t[/tex]
 
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The expression for average velocity is correct, but you calculated the time wrong. 1 hour equals 3600 seconds. So, 0.2 hours equals 720 seconds.
 
Ah, thank you.

I corrected my mistake and fixed my answers and they're correct.
 

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