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UNIT 1: 1-D KINEMATICS
Lesson 4:
Graphical Analysis of Accelerating Motion
CENTRE HIGH: PHYSICS 20
Reading Segment #1:
Average and Instantaneous Velocity on Position-Time Graphs
To prepare for this section, please read:
Unit 1: p.14
D1. Position-Time Graphs (Accelerated Motion)
Recall, when the motion of an object is uniform,
- the velocity is constant
- since velocity is the slope of a position-time graph,
the graph is a straight line (constant slope).
But now, the velocity is going to change.
As a result, the line is no longer going to be straight.
Consider the following position-time graph
for an accelerating object: (Forward is positive)
d (m)
t (s)
Can you describe the motion?
d (m)
Zero slope
t (s)
At the start, the slope is zero.
Thus, the object starts from rest.
d (m)
Slopes are increasing
t (s)
But, over time, the slope increases.
Thus, the object's velocity increases, which in this case
means that the object is speeding up.
d (m)
Slopes are increasing
t (s)
Since the velocity is changing,
the object is accelerating.
For non-linear position-time graphs, there are two types
of velocities (i.e. slopes) that could be calculated:
1. Average velocity
- from one time to another
2. Instantaneous velocity
- at one point in time
1. Average velocity
To find the average velocity from t = t1 to t = t2,
Find the initial position (at t1) and
the final position (at t2) on the graph
Draw a straight line from the two positions
(called a secant line)
Find the slope of this secant line
e.g. How would we find the average velocity from t = 3.0 s
to t = 5.0 s on the graph below?
d (m)
3.0 5.0 t (s)
d (m)
3.0 5.0 t (s)
First, find the initial and final position on the line.
d (m)
secant line
3.0 5.0 t (s)
Draw a straight line between the two points.
This is called the secant line.
d (m)
slope = vavg
3.0 5.0 t (s)
Find the slope of the secant line.
The slope is the average velocity between 3.0 and 5.0 s.
2. Instantaneous velocity
To find the instantaneous velocity at t = t1 :
Find the position (at t1) on the graph
Draw a tangent line at this point
Choose 2 points on the tangent line and find its slope
e.g. How would we find the instantaneous velocity at t = 4.0 s ?
d (m)
4.0 t (s)
d (m)
4.0 t (s)
First, find the position at t = 4.0 s on the line.
d (m)
tangent line
4.0 t (s)
Next, draw a tangent line at this point.
A tangent line touches the curve only at this point,
and if done correctly, it should show the slope at that point.
d (m)
slope of tangent
= velocity
4.0 t (s)
Finally, locate two points on the tangent line
and find its slope.
The slope is the instantaneous velocity.
Ex. 1 For the graph below, find: (North is positive)
a) the average velocity from 0.25 s to 5.0 s
b) the instantaneous velocity at 1.0 s
02468101214161820222426283032343638404244464850
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (s)
Po
siti
on
(m
)
a) Locate the positions at t = 0.25 s and t = 5.0 s
02468101214161820222426283032343638404244464850
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (s)
Po
siti
on
(m
)
secant line
a) Draw a straight line between the two points.
This the secant line.
02468101214161820222426283032343638404244464850
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (s)
Po
siti
on
(m
)
02468101214161820222426283032343638404244464850
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (s)
Po
siti
on
(m
)
(5.0 s, 50.0 m)
secant line
(0.25 s, 11.0 m)
a) Choose two points on the secant line and find the slope.
a) Average velocity between 0.25 s and 5.0 s:
vavg = slope of secant line
= y2 - y1 = (50.0 m) - (11.0 m)
x2 - x1 (5.0 s) - (0.25 s)
= 8.2 m/s North
Be sure to include the direction for velocity.
b) Locate the position at t = 1.0 s
02468101214161820222426283032343638404244464850
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (s)
Po
siti
on
(m
)
tangent
line
b) Draw the tangent line at this point.
It should touch the curve only at this point.
02468101214161820222426283032343638404244464850
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (s)
Po
siti
on
(m
)
(3.25 s, 48.0 m)
tangent
line
(1.0 s, 22.0 m)
b) Find two points on the tangent line. Calculate the slope.
02468101214161820222426283032343638404244464850
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (s)
Po
siti
on
(m
)
b) Instantaneous velocity between 1.0 s:
v1.0 = slope of tangent line
= y2 - y1 = (48.0 m) - (22.0 m)
x2 - x1 (3.25 s) - (1.0 s)
= 12 m/s North
Be sure to include the direction for velocity.
Practice Problems
Try these problems in the Physics 20 Workbook:
Unit 1 p. 16 #1
Reading Segment #2:
Velocity-Time Graphs
To prepare for this section, please read:
Unit 1: p.15
D2. Velocity-Time Graphs (Accelerated Motion)
There are two major calculations we find for velocity-time
graphs:
1. Acceleration
- this will be the slope of a v-t graph
2. Displacement
- this will be the area "under" the v-t graph
1. Acceleration on a velocity-time graph
The slope of a v-t graph is the acceleration
Thus, if the object has uniform acceleration:
- the acceleration is constant
- the slope is constant on the v-t graph
- this produces a straight line
e.g. Ref: Forward is positive
Backward is negative
If the graph has a positive slope (i.e. a positive acceleration),
it means the object's velocity is increasing.
v v
t
t
Ref: Forward is positive
Backward is negative
If the graph has a negative slope (i.e. a negative acceleration),
it means the object's velocity is decreasing.
v v
t
t
Ref: Forward is positive
Backward is negative
If the graph has a zero slope (i.e. a zero acceleration),
it means the object has a constant velocity.
v v
t
t
Ex. 2 For the graph below, find the acceleration at t = 3.0 s.
(East is positive)
Velocity 8.0
( 103 m/s)
2.0
4.0 t (s)
Velocity 8.0
( 103 m/s)
2.0
4.0 t (s)
Acceleration is the slope of a v-t graph.
Since this is a straight line, you can choose any 2 points
on the line to calculate the slope.
Velocity 8.0 (0 s, 8.0 103 m/s)
( 103 m/s)
(4.0 s, 2.0 103 m/s)
2.0
4.0 t (s)
Choose 2 points on the line.
Acceleration at 3.0 s:
a3.0 = slope of line
= y2 - y1 = (2.0 103 m/s) - (8.0 103 m/s)
x2 - x1 (3.0 s) - (0 s)
= -1.5 103 m/s2
= 1.5 103 m/s2 West
Be sure to include the direction for acceleration.
Ex. 3 Describe the motion showed by the following
velocity-time graph: (Forward is positive)
v (m/s)
10
t (s)
-6
v (m/s) Forward is positive
10 A
t (s)
-6
At the very start of the motion (A), the object is moving
forward at 10 m/s.
v (m/s) Forward is positive
10 A
B t (s)
-6
Along AB, the object is slowing down at a constant acceleration (constant slope), from 10 m/s to 0.
It is still moving forward, since the velocity is positive.
At B, the object is at rest.
v (m/s) Forward is positive
10 A
B t (s)
-6 C
Along BC, the object is speeding up at a constant acceleration (constant slope), from 0 to 6 m/s.
It is now moving backward, since the velocity is negative.
At C, the object is at moving backwards at 6 m/s.
v (m/s) Forward is positive
10 A
B t (s)
-6 C D
Along CD, the object is at a constant velocity of -6.0 m/s.
Acceleration (slope) is zero.
At D, the object is at moving backwards at 6 m/s.
v (m/s) Forward is positive
10 A
B E t (s)
-6 C D
Along DE, the object is slowing down at a constant
acceleration (slope), from 6 m/s to zero.
It is moving backwards, since the velocities are negative.
At E, the object has come to rest.
2. Displacement on a velocity-time graph
The area between the line and the t-axis on a
v-t graph is the displacement
Ref: Forward is positive
Backward is negative
e.g. When the line is above the t-axis,
displacement (area) is positive.
i.e. A forward displacement
v v
t t
Ref: Forward is positive
Backward is negative
e.g. When the line is below the t-axis,
displacement (area) is negative.
i.e. A backward displacement
v v
t t
Ex. 4 Calculate the displacement for the first 8.0 seconds.
(North is positive)
v (m/s)
33
14
8.0 t (s)
v (m/s)
33
14
8.0 t (s)
Displacement is the area between the line
and the t-axis
v (m/s)
33
1
14
2
8.0 t (s)
To calculate the area, find the area of the triangle (1)
and the rectangle (2).
v (m/s)
33
1 19 m/s
14
2 14 m/s
8.0 s 8.0 t (s)
Show the dimensions on the diagram.
Displacement for the first 8.0 seconds:
A1 = Area of triangle
= 0.5 b h
= 0.5 (8.0 s) (19 m/s) = 76 m
A2 = Area of rectangle
= L w
= (8.0 s) (14 m/s) = 112 m
Thus, d = A1 + A2
= 1.9 102 m North
Practice Problems
Try these problems in the Physics 20 Workbook:
Unit 1 pp. 16 - 17 #2 - 5