# Graphs

```Topic
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Displacement
Vectors
Kinematics
Graphs
Energy
Power
Springs
Field of Vision
Colors
Concave mirrors
Convex mirrors
Refraction
Lenses
Optical Power
Slides Minutes
9
27
13
39
13
39
10
30
10
30
5
15
4
12
3
9
7
21
3
9
7
21
4
12
5
15
10
30
6
18
Distance-Time
Graphs
Slope = m / s = velocity
Area = m x s = ms = nothing
8
s (m)
Distance-Time Graph
7
6
Slope = velocity
Area = nothing
5
4
3
2
1
t(s)
0
1
a)
b)
c)
d)
Example
At
At
At
At
2
3
4
5
6
7
8
9
10 11 11
12 13 14
t = 2 s velocity = 1 m/s (slope = 1)
t = 5 s velocity = 0 (slope = 0)
t = 9 s velocity = -1 m/s (slope = -1)
t = 12 s velocity = 0 (slope = 0)
Click to continue
Distance –vs- Time graph
In a distance versus time graph, the slope represents the velocity.
And the area under the curve, does not represent anything.
80
s (m)
70
60
50
40
30
20
Area = nothing
10
t (s)
0
1
2
3
4
5
6
7
8
9
10 11 11
12 13 14
To find the velocity at any point, find the slope at that point.
Click
Velocity-Time
Graphs
In a velocity versus time graph, the slope represents the acceleration.
And the area under the curve, represents the distance traveled.
80
v (m/s)
70
60
50
40
30
20
Area = distance traveled
10
t (s)
0
1
2
3
4
5
6
7
8
9
10 11 11
12 13 14
To find the acceleration at any point, find the slope at that point.
To find the distance traveled between any two points, find the area
under the curve between those two points.
Click
8
v (m/s)
7
Velocity-Time Graph
6
Slope = acceleration
Area = distance
5
The graph on the left
illustrates the velocity-time
curve of a vehicle.
4
Find the total distance
traveled by the vehicle.
3
2
C
1
A
B
D
t(s)
0
1
2
3
4
5
6
7
8
9
10 11 11
12 13 14
Area of sector A = (3 m/s x 3 s)/2 = 4.5 m
Area of sector B = 3 m/s x 5 s = 15 m
There
arewe
a number
ways
oftime,
finding
the
total
area
under
curve.
Since
First
velocity
find
is plotted
theCof
area
versus
of each
segment
we
know
that
then
slope
up aall
represents
the
Area
of sector
= (3
m/s
x 2 s)/2
=2
mand
Inacceleration
the
example
above,
one
way
is
to
divide
the
total
into
To
the
total distance,
we need
to the
find
the
total
area
under
thefour
curve.
segments
obtain
the
total
area
and
thus
thearea
thefind
oftothe
vehicle
and
area
under
the
curve
represents
Area of (segment-A,
sector D = 1 segment-B,
m/s x 6 s = 6segment-C
m
sections
and segment-D).
the distance traveled by the vehicle.
Distance traveled = A + B + C + D = 4.5 m + 15 m + 2 m + 6 m = 27.5 m
Example-1
Click to continue
8
7
v (m/s)
Velocity-Time Graph
6
The graph on the left
illustrates the velocity-time
curve of a vehicle.
Slope = acceleration
Area = distance
5
4
Find the distance traveled by
the vehicle for the first 5 s.
3
2
1
A
B
t(s)
0
1
2
3
4
5
6
7
8
9
10 11 11
12 13 14
Step-1: Area of sector A = (3 m/s x 3 s)/2 = 4.5 m
Note
that
the
first
5for
seconds
is5 from
t =curve
0 tooft two
= 5 s.
Thus,Since
we need
the to
distance
find
the
distance
the firstunder
s consists
the
from
segments,
t = 0 to t = 5 s.
Step-2: Area of sector A = (3 m/s x 2 s) = 6 m
we will find the area in two steps. Step-1 is from t = 0 to t = 3 s (segment A)
and step-2 is from t = 3 s to t = 5 s (segment B).
Distance traveled = A + B = 4.5 m + 8 m = 10.5 m
Example-2
Click to continue
8
7
v (m/s)
Velocity-Time Graph
6
The graph on the left
illustrates the velocity-time
curve of a vehicle.
Slope = acceleration
Area = distance
5
Find the distance traveled by
the vehicle between t = 4 s
and t = 7 s.
4
3
2
1
t(s)
0
1
2
3
4
5
6
7
8
9
10 11 11
12 13 14
Note that
Area
the under
area under
the curve
the curve
= 3 m/s
for xthis
3 sproblem
= 9 m is from t = 4 s to t = 7 s
Thus, all we need to do is find the area of the rectangle.
and consists of only one section.
Distance traveled = 9 m
Example-3
8
7
v (m/s)
Velocity-Time Graph
6
The graph on the left
illustrates the velocity-time
curve of a vehicle.
Slope = acceleration
Area = distance
5
Find the distance traveled by
the vehicle during the last
5 seconds.
4
3
2
1
t(s)
0
1
2
3
4
5
6
7
8
9
10 11 11
12 13 14
1 m/s by 1 s
1 m/s by 5 s
The area under the curve for this problem consists of two sections,
Note that the area under the curve for this problem is from t = 9 s to t = 14 s.
a small
Area of triangle
= (1triangle
m/s x 1and
s)/2a =rectangle.
0.5 m
Area of rectangle = (1 m/s x 5 s) = 5 m
Distance traveled = 0.5 m + 5 m = 5.5 m
Example-4
Graphs Slide: 4. 1
During the course of a laboratory experiment, a team of students
obtained the graph on the right representing the force exerted as
a function of the acceleration of a cart.
Calculate the mass of the cart.
A)
3.0 kg
B)
2.0 kg
C)
1.0 kg
D)
0.50 kg
E)
0.25 kg
Note that the slope
represents mass.
Graphs Slide:
Reminder
The area under the curve
is the distance travelled.
4. 2
200 m
250 m
200 m
+ 100 m
Answer: 350 m – 200 m = 150 m
350 m
Click
Graphs Slide: 4. 3
Illustrated below is the velocity versus time graph of a particle.
NOTE: The area under the curve represents distance.
250 000
50 000
200 000
How far has the particle travelled in 5 minutes?
A) 200 000 m
B) 250 000 m
C) 300 000 m
D) 350 000 m
Convert to seconds
5 min = 5 x 60 s = 300 s
Graphs Slide: 4. 4
Consider the position versus time graph below.
Which one of the following statements best describes the
motion illustrated by the above graph.
A) Increasing velocity, constant velocity, increasing velocity
B) Increasing velocity, zero velocity, increasing velocity
C) Constant velocity, constant velocity, constant velocity
D) Constant velocity, zero velocity, constant velocity
Click
Zero acceleration
Positive
acceleration
Graphs Slide:
4. 5
The velocity-time graph on the
left represents the motion of a
car during a 6 s interval of time.
Negative acceleration
A)
B)
Which of these
graphs represents
the acceleration
of the car?
C)
D)
Click
Graphs Slide:
4. 6
Fastest velocity here thus
maximum KE at this point.
This point shows that the car
has zero KE which means it
has zero velocity at point I
which is incorrect.
Click
Graphs Slide: 4. 7
Velocity = 750 m/25 s = 30 m/s
Area = velocity = 50 m/s
Velocity = 600 m/25 s = 24 m/s
Velocity = 25 m/s
Click
Slide:
4. 8
This means
second
segment
has a
negative
This the
means
the first
segment
has
to be aacceleration
straight line (slope).
(a=0).
Click
Graphs Slide: 4. 9
Since the distance between
dots is increasing, the object
is accelerating.
Zero velocity
Constant
forward
velocity
Constant
reverse
velocity
Click
Graphs Slide: 4. 10
Carlo Martini owns a Ferrari. Carlo performed a speed test on
his car and plotted the graph below.
Note that the slope represents
acceleration.
Knowing that Carlo’s Ferrari has a mass of 1288 kg, calculate
the net force of the car based on the above graph.
Click
Graphs Slide: 4. 13
Both a car and a truck drive off at the same time and in the same
direction. The graphs below illustrate their movement.
Area to x-axis = distance
Slope = Average velocity = 10 m/s
Determine how far apart the two vehicles are after 30 seconds.
Step-1 Distance of car = area under the curve = 275 m
Step-2 Distance of car = VAt = (10 m/s)(30 s) = 300 m
Step-3 Distance apart = 300 m – 275 m = 25 m
Click
… and good luck!
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