5. t(s)
Example
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Distance-Time Graph
Slope = velocity
Area = nothing
s (m)
Click to continue
a) At t = 2 s velocity = 1 m/s (slope = 1)
b) At t = 5 s velocity = 0 (slope = 0)
c) At t = 9 s velocity = -1 m/s (slope = -1)
d) At t = 12 s velocity = 0 (slope = 0)
Slope = m / s = velocity
Area = m x s = ms = nothing
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s (m)
t (s)
In a distance versus time graph, the slope represents the velocity.
And the area under the curve, does not represent anything.
Area = nothing
To find the velocity at any point, find the slope at that point.
Click
Distance –vs- Time graph
8. 0
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v (m/s)
t (s)
In a velocity versus time graph, the slope represents the acceleration.
And the area under the curve, represents the distance traveled.
Area = distance traveled
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
9. Area of sector D = 1 m/s x 6 s = 6 m
t(s)
A B
C
D
Distance traveled = A + B + C + D = 4.5 m + 15 m + 2 m + 6 m = 27.5 m
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
Area of sector C = (3 m/s x 2 s)/2 = 2 m
Example-1
To find the total distance, we need to find the total area under the curve.
There are a number of ways of finding the total area under a curve.
In the example above, one way is to divide the total area into four
sections (segment-A, segment-B, segment-C and segment-D).
First we find the area of each segment and then add up all the
segments to obtain the total area and thus the answer.
Since velocity is plotted versus time, we know that the slope represents
the acceleration of the vehicle and the area under the curve represents
the distance traveled by the vehicle.
The graph on the left
illustrates the velocity-time
curve of a vehicle.
Find the total distance
traveled by the vehicle.
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Velocity-Time Graph
Slope = acceleration
Area = distance
v (m/s)
Click to continue
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v (m/s)
t(s)
Example-2
Velocity-Time Graph
Slope = acceleration
Area = distance
Note that the first 5 seconds is from t = 0 to t = 5 s.
Thus, we need to find the distance under the curve from t = 0 to t = 5 s.
Since the distance for the first 5 s consists of two segments,
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).
A B
Step-1: Area of sector A = (3 m/s x 3 s)/2 = 4.5 m
Step-2: Area of sector A = (3 m/s x 2 s) = 6 m
Distance traveled = A + B = 4.5 m + 8 m = 10.5 m
The graph on the left
illustrates the velocity-time
curve of a vehicle.
Find the distance traveled by
the vehicle for the first 5 s.
Answer
Click to continue
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v (m/s)
t(s)
Example-3
Velocity-Time Graph
Slope = acceleration
Area = distance
Note that the area under the curve for this problem is from t = 4 s to t = 7 s
and consists of only one section.
Thus, all we need to do is find the area of the rectangle.
Area under the curve = 3 m/s x 3 s = 9 m
Distance traveled = 9 m
The graph on the left
illustrates the velocity-time
curve of a vehicle.
Find the distance traveled by
the vehicle between t = 4 s
and t = 7 s.
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v (m/s)
t(s)
Example-4
Velocity-Time Graph
Slope = acceleration
Area = distance
Note that the area under the curve for this problem is from t = 9 s to t = 14 s.
The area under the curve for this problem consists of two sections,
a small triangle and a rectangle.
Area of triangle = (1 m/s x 1 s)/2 = 0.5 m
Distance traveled = 0.5 m + 5 m = 5.5 m
The graph on the left
illustrates the velocity-time
curve of a vehicle.
Find the distance traveled by
the vehicle during the last
5 seconds.
1 m/s by 1 s
Area of rectangle = (1 m/s x 5 s) = 5 m
1 m/s by 5 s
Answer
13.
14. 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: 4. 1
15. Reminder
The area under the curve
is the distance travelled.
200 m
250 m
+ 100 m
350 m
200 m
Answer: 350 m – 200 m = 150 m
Click
Graphs Slide: 4. 2
16. Illustrated below is the velocity versus time graph of a particle.
A) 200 000 m
B) 250 000 m
C) 300 000 m
D) 350 000 m
How far has the particle travelled in 5 minutes?
Convert to seconds
NOTE: The area under the curve represents distance.
5 min = 5 x 60 s = 300 s
50 000 200 000
250 000
Graphs Slide: 4. 3
17. Consider the position versus time graph below.
Click
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
Which one of the following statements best describes the
motion illustrated by the above graph.
Graphs Slide: 4. 4
18. Click
Which of these
graphs represents
the acceleration
of the car?
The velocity-time graph on the
left represents the motion of a
car during a 6 s interval of time.
A)
B)
C)
D)
Positive
acceleration
Zero acceleration
Negative acceleration
Graphs Slide: 4. 5
19. 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. 6
20. Velocity = 25 m/s
Velocity = 750 m/25 s = 30 m/s
Area = velocity = 50 m/s
Velocity = 600 m/25 s = 24 m/s
Click
Graphs Slide: 4. 7
21. This means the first segment has to be a straight line (a=0).
This means the second segment has a negative acceleration (slope).
Click
4. 8
Slide:
22. Since the distance between
dots is increasing, the object
is accelerating.
Constant
forward
velocity
Zero velocity
Constant
reverse
velocity
Click
Graphs Slide: 4. 9
23. Carlo Martini owns a Ferrari. Carlo performed a speed test on
his car and plotted the graph below.
Click
Knowing that Carlo’s Ferrari has a mass of 1288 kg, calculate
the net force of the car based on the above graph.
Note that the slope represents
acceleration.
Graphs Slide: 4. 10
24. Click
Both a car and a truck drive off at the same time and in the same
direction. The graphs below illustrate their movement.
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
Area to x-axis = distance Slope = Average velocity = 10 m/s
Step-3 Distance apart = 300 m – 275 m = 25 m
Graphs Slide: 4. 13
