1. Chapter 2 – Motion in one dimension Time and Position
PHY 113 C General Physics I
11 AM – 12:15 PM TR Olin 101
Plan for Lecture 2:
Chapter 2 – Motion in one dimension
Time and Position
Time and Velocity
Time and Acceleration
Special relationships for constant acceleration
Problems 1.1,1.6,1.10,1.11
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PHY 113 C Fall Lecture 2

2. Some updates/announcements
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PHY 113 C Fall Lecture 2

3. Webassign Experiences so far
iclicker exercises:
Webassign Experiences so far
Have not tried it
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Can login
Have logged in and have completed one or more example problems.
Textbook Experiences
Have no textbook (yet)
Have complete physical textbook
Have electronic version of textbook
Textbook is on order
Other
8/29/2013
PHY 113 C Fall Lecture 2

4. Comment on Webassign #1 problem:
7. A pet lamb grows rapidly, with its mass proportional to the cube of its length. When the lamb's length changes by 16%, its mass increases by 16.8 kg. Find the lamb's mass at the end of this process.
Problem solving steps
Visualize problem – labeling variables
Determine which basic physical principle(s) apply
Write down the appropriate equations using the variables defined in step 1.
Check whether you have the correct amount of information to solve the problem (same number of knowns and unknowns).
Solve the equations.
Check whether your answer makes sense (units, order of magnitude, etc.).
8/29/2013
PHY 113 C Fall Lecture 2

5. 7. A pet lamb grows rapidly, with its mass proportional to the cube of its length. When the lamb's length changes by 16%, its mass increases by 16.8 kg. Find the lamb's mass at the end of this process. L
8/29/2013
PHY 113 C Fall Lecture 2

6. i>clicker exercise: What did you learn from this problem?
It is a bad idea to have a pet lamb
This was a very hard question
I hope this problem will not be on an exam
My physics instructor is VERY mean
All of the above.
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PHY 113 C Fall Lecture 2

7. Motion in one-dimension
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PHY 113 C Fall Lecture 2

8. Graphical representation of position (displacement) x(t)
t (s)
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PHY 113 C Fall Lecture 2

9. Comment:
Your text mentions the notion of a “scalar quantity” in contrast to a “vector quantity” which will be introduced in Chapter In most contexts, a scalar quantity – like one-dimensional distance or displacement can be positive or negative.
Another comment:
Your text describes the time rate of change of displacement as “velocity” which, in one-dimension is a signed scalar quantity. In general “speed” is the magnitude of velocity – a positive scalar quantity.
8/29/2013
PHY 113 C Fall Lecture 2

10. Graphical representation of position (displacement): x(t)  time rate of change of displacement = velocity: v(t)
t (s)
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PHY 113 C Fall Lecture 2

11. Instantaneous velocityvB vC vA vD vE vF
t (s)
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PHY 113 C Fall Lecture 2

12. Demonstration of tangent line limit
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PHY 113 C Fall Lecture 2

13. Average velocity versus instantaneous velocity
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PHY 113 C Fall Lecture 2

14. Average velocity iclicker exercise: This results is: Exact Approximate
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PHY 113 C Fall Lecture 2

15. Webassign Example
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PHY 113 C Fall Lecture 2

16. Instantaneous velocity using calculusx v
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PHY 113 C Fall Lecture 2

17. Anti-derivative relationshipx t
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PHY 113 C Fall Lecture 2

18. Velocity
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PHY 113 C Fall Lecture 2

19. Acceleration
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PHY 113 C Fall Lecture 2

20. Rate of acceleration iclicker exercise
How many derivatives of position are useful for describing motion:
1 (dx/dt)
2 (d2x/dt2)
3 (dx3/dt3)
4 (dx4/dt4) 
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PHY 113 C Fall Lecture 2

21. Anti-derivative relationship
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PHY 113 C Fall Lecture 2

22. Examples
v(t)
x(t)
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PHY 113 C Fall Lecture 2

23. Examples
v(t)
x(t)
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PHY 113 C Fall Lecture 2

24. Summary
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PHY 113 C Fall Lecture 2

25. Another example x(t) v(t)
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PHY 113 C Fall Lecture 2

26. From webassign:
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PHY 113 C Fall Lecture 2

27. Another example (m/s) (s)
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PHY 113 C Fall Lecture 2

28. Another example -- continued
(m/s)
(s)
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PHY 113 C Fall Lecture 2

29. Special relationships between t,x,v,a for constant a:
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PHY 113 C Fall Lecture 2

30. Special relationships between t,x,v,a for constant a:
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PHY 113 C Fall Lecture 2

31. Special case: constant velocity due to earth’s gravity
In this case, the “one” dimension is the vertical direction with “up” chosen as positive:
a = -g = -9.8 m/s2
y(t) = y0 + v0t – ½ gt2
y0 = v0 = 20 m/s
At what time tm will the ball hit the ground
ym = -50m ?
Solve: ym = y0 + v0tm – ½ gtm2
quadratic equation
physical solution: tm = 5.83 s
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PHY 113 C Fall Lecture 2

32. Helpful mathematical relationships (see Appendix B of your text)
8/29/2013
PHY 113 C Fall Lecture 2