Reading the position - time graph:

In summary, the physics professor walks home from campus via the sidewalk, and it starts to rain. Her distance from her house as a function of time is shown in the figure below. At which of the labeled points is her velocity zero?
  • #1
Edwardo_Elric
101
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
 
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  • #2
Was that graph given to you in the question, or have you drawn it from a previous part?
 
  • #3
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)
 
  • #4
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
 
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  • #5
No Problem
 
  • #6
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.
 

FAQ: Reading the position - time graph:

What is a position-time graph?

A position-time graph is a visual representation of an object's position as it changes over time. The horizontal axis represents time and the vertical axis represents position.

How do you interpret a position-time graph?

To interpret a position-time graph, you can look at the slope of the line, which represents the object's velocity. A steeper slope indicates a greater velocity, while a flatter slope indicates a slower velocity. You can also look at the x-intercept, which represents the object's initial position, and the y-intercept, which represents the object's displacement.

What does a straight line on a position-time graph mean?

A straight line on a position-time graph means that the object is moving at a constant velocity. The slope of the line represents the magnitude and direction of the velocity.

How can you determine an object's acceleration from a position-time graph?

An object's acceleration can be determined by looking at the curvature of the line on a position-time graph. A curved line indicates a changing velocity, which means the object is accelerating. The steeper the curvature, the greater the acceleration.

Can you calculate an object's displacement from a position-time graph?

Yes, an object's displacement can be calculated by finding the difference between the final and initial positions on a position-time graph. This is represented by the change in the y-values or the area under the curve.

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