the graphs… game. today we are going to describe motion using motion graphs... first, let’s look...
TRANSCRIPT
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Today we are going to describe motion using motion graphs . . .• First, let’s look at
constant position, or no motion
• The object would have the same position over a period of time on the graph.
• The y-intercept shows how far the object is from a reference point.
d
t
No Motion or Constant Position
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a
t
No Motion or Constant Position
v
t
No Motion or Constant Position
If we were to plot that zero velocity vs. time, it would look like -
And the resulting zero acceleration vs. time graph would look like -
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Consider a car moving with a constant positive (rightward) velocity of 10 m/s
What would a Position vs Time graph of this motion look like?
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The slope of the line on a position-time graph is the velocity of the object.
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And the slope is…
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Example A car moves in a straight line and its odometer readings are plotted against time.a. Find the speed of the car at point A.
t
x
08
04
= 0.5 m/s
A
B
Same for point B
b. What is the car's average speed?
= 0.5 m/s v
c. What is its acceleration?
a = 0 m/s2
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v
t
Constant Positive Velocity
d
t
Constant Positive Velocity
So the position vs time, or d vs t graph would look like -
And the velocity vs time graph would look like -
![Page 9: The graphs… Game. Today we are going to describe motion using motion graphs... First, let’s look at constant position, or no motion The object would have](https://reader030.vdocuments.net/reader030/viewer/2022032707/56649e4d5503460f94b4406d/html5/thumbnails/9.jpg)
And once again since the acceleration is zero, the acceleration vs time graph would look like -
a
t
Constant Positive Velocity
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A constant positive velocity means an object is moving away from a fixed reference point.
A constant negative velocity means an object is moving towards a fixed reference point.
d
t
Constant Negative Velocity
So the distance vs time graph would look like -
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vt
Constant Negative Velocity
a
t
Constant Negative Velocity
The slope would be a negative number. The v vs t plot would be -
And once again since acceleration is zero, the plot would be -
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The red car is moving faster, so the slope the red line is larger representing a larger velocity.
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Let’s compare the motions shown here
Compare the velocity of the red line to the…
Green line Blue line
Black line
Orange line
And this trip shows?
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Now consider a car moving with a changing, rightward(+) velocity, i.e. it is moving rightward and speeding up or accelerating.
And the graph’s shape is?
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The instantaneous velocity of an object at a certain time is the slope of the position (displacement) versus time graph at that time. It can be positive, negative, or zero.
What shows the instantaneous speed in the car?
The speedometer!
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Looking back to the car with changing velocity -
What would its velocity vs. time graph look like?
First, we need to calculate the velocity:
Time Velocity0 0
1 sec 2 m/s
2 sec 4 m/s
3 sec 6 m/s
4 sec 8 m/s
5 sec 10 m/s
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A plot of the velocity vs. time would look like -
and the slope (ΔV/ΔT) is the acceleration of the object.
ΔVΔT
= 4 m/s2 sec
= 2 m/s2
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v
t
Constant Positive Acceleration
d
t
Constant Positive Acceleration
So for a constantly accelerating object, the d vs t graph is -
Plotting the velocity over time would look like -
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a
t
Constant Positive
Acceleration
Since the v vs t graph has a slope, that would give the acceleration and the graph would look like -
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FYI: Constant Negative Acceleration graphs would look like -
d
t
Constant Negative Acceleration or Deceleration
v
t
Constant Negative
Acceleration
at
Constant Negative
Acceleration