Reading the position - time graph:

Click For Summary
SUMMARY

The discussion centers on analyzing a distance-time graph depicting a physics professor's walk to campus and back. Key points include that her velocity is zero at the turning point, constant and positive while moving away from home, constant and negative while returning, increasing in magnitude between points I and II, and decreasing in magnitude between points III and IV. The participants clarify that constant velocity applies over intervals, not at single points, and emphasize the importance of graph curvature in determining velocity changes.

PREREQUISITES
  • Understanding of distance-time graphs
  • Knowledge of velocity concepts in physics
  • Familiarity with terms like "convex up" and "convex down"
  • Ability to interpret graphical data
NEXT STEPS
  • Study the properties of distance-time graphs in physics
  • Learn about instantaneous vs. average velocity
  • Explore the implications of graph curvature on motion
  • Review concepts of acceleration and its graphical representation
USEFUL FOR

Students of physics, educators teaching motion concepts, and anyone interested in understanding the relationship between distance, time, and velocity through graphical analysis.

Edwardo_Elric
Messages
101
Reaction score
0

Homework Statement


A physics proffessor leaves her house and walks along the sidewalk toward campus. After 5 minutes, it starts to rain, and she returns home. Her distance from her house as a function of time is shown in the figure below.(ignore the paintbrush just assume that its a curved line)
At which of the labeled points is her velocity...
a.) zero?
b.) constant and positive?
c.) constant and negative?
d.) increasing in magnitude?
e.) decreasing in magnitude?

Homework Equations


none


The Attempt at a Solution


a.) zero?
none... i think zero is from the end after IV

b.) constant and positive?
none?

c.) constant and negative?
point V

d.) increasing in magnitude?
from points I - II

e.) decreasing in magnitude?
from points III - IV
Distancetimegraph-1.jpg
 
Last edited:
Physics news on Phys.org
Was that graph given to you in the question, or have you drawn it from a previous part?
 
You should be right (I think)
0 velocity exists at 0 distance (say displacement/deltaT)...velocity is zero seemingly only at limt = 0 and when her journey ends. 0 velocity also exists when limit of the point is a tangent line parallel to the x-axis...so...
a)Not an Optional Point
b)No Point
c)Point V
d)Point I and Point II
e)Point III and Point IV
-----------------Same Answers as Yours----------------------
(I'm just an amateaur at the moment, so I do not know if this will be of any help)
 
i have no scanner but i think this is likely similar to the graph in my book
and the brush ... its represented as curves
and thanks Gear300 for your opinions
 
Last edited:
No Problem
 
Unfortunately, I'm not sure Gear300's answers are very helpful- he seems to be thinking of average velocity.

Yes, you are correct that as long as the graph is going upward, velocity is positive, as long as it is going downward it is negative. At the point where she "turns back", changing from going away from her home (positive velocity) to going toward her home (negative velocity) here velocity is 0. Unfortunately, on your graph, that is not one of the labled points!

I'm not happy with the wording of B or C: it makes no sense to say that the velocity is "constant" at a single point- "constant" velocity can only apply over a time interval. On a distance-time graph, a velocity will be constant where the graph is a straight line. Yes, the velocity is constant (and negative) on the interval containing V. The velocity is increasing where the graph is "convex up" and decreasing where it is "convex down"- it looks likethose occur at I and III respectively.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
Replies
3
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
22
Views
4K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K