Uniformly accelerated motion for Grade 9

1. UNIFORMLY ACCELERATED
MOTION
PREPARED BY:
MADEN ROBIE JANE C. CRISOSTOMO
INTERN

2. Activity 1.
Word Association
Direction: Think of word/s or situations that can be
associated from the words below.
2

3. DISTANCE
▷ the actual length
covered by an object
▷ It is a scalar quantity
meaning it only has a
numerical value
3

4. DISPLACEMENT
▷ the straight line or
short cut from the
initial position to the
final position with
direction
▷ It is a vector quantity
which means it both
has a numerical value
and direction
4

5. SPEED
▷ Speed is how fast you
are traveling
▷ eg. riding a car with a
speed of 20 m/s
5

6. VELOCITY
▷ the speed in a given
direction.
▷ eg. riding a car with a
speed of 20 m/s going
North
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7. ACCELERATION
▷ the change in velocity
over time
▷ means, that there is a
change or variation of
distance at a given
time
▷ eg. the car’s
accelerating at 20
m/s2. 7
20m/s2

8. Assume the you are riding a vehicle and
observed the speedometer readings below.
8
ELAPSED TIME SPEEDOMETER
READING
After 1 minute 10 m/s
After 2 minutes 20 m/s
After 3 minutes 30 m/s

9. Uniformly Accelerated
Motion
▷ is motion of an object where the
acceleration is constant.
9

10. 10
2 Types of UAM
1. Horizontal Dimension
 x-axis
 Motion straight in the
ground
2. Vertical Dimension
 y-axis
 Also called free fall
 Motion upward and
downward

11. Countdown to 5
5 variables
4 formulas
3 able to find 3 variables
2 you can obtain 2 remaining variables
1 happy physics student
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12. 5 Variables
12
VARIABLE UNIT
a = acceleration m/s2
vi = initial velocity m/s
vf = final velocity m/s
t = time s
d = distance/displacement m

13. 4 Formulas
13
1. 𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡
2. 𝑑 = 𝑡
𝑣𝑓− 𝑣𝑖
2
3. 𝑑 = 𝑣𝑖𝑡 +
1
2
𝑎𝑡2
4. vf2 = vi2 + 2ad

14. Horizontal Dimension
Problem Solving
14

15. 1. An airplane from rest accelerates on a
runway at 5.50 𝑚/𝑠2
for 20.25 s. until it finally
takes off the ground. What is the distance
covered before take-off?
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Given:
a = 5.50 𝑚/𝑠2
t = 20.25 s
vi = 0 m/s
d =?
Formula:
𝑑 = 𝑣𝑖𝑡 +
1
2
𝑎𝑡2

16. 16
Given:
a = 5.50 𝑚/𝑠2
t = 20.25 s
vi = 0 m/s
d =?
Formula:
𝑑 = 𝑣𝑖𝑡 +
1
2
𝑎𝑡2
Solution:
𝑑 = 𝑣𝑖𝑡 +
1
2
𝑎𝑡2
d = (0 m/s) (20.25 s) + (0.5) (5.50 m/s2)
(20.25s)2
d = (0 m/s) (20.25 s) + (0.5) (5.50 m/s2)
(20.25s)2
d = (0.5) (5.50 m/s2) (410.0625s2)
d = (0.5) (2, 255.34375m)
d = 1, 127.67 m or 1, 127.70m
d = 1, 127.70m

17. 2. A car at rest accelerated at 8m/s2 for 10 s.
A. What is the position of the car at the
end of 10 seconds?
B. What is the velocity of the car at the
end of 10 seconds?
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18. 18
A. What is the position of the car at the end
of 10 seconds?
Given:
a = 8m/s2
t = 10s
vi = 0m/s
d = ?
Formula:
𝑑 = 𝑣𝑖𝑡 +
1
2
𝑎𝑡2

19. 19
Formula:
𝑑 = 𝑣𝑖𝑡 +
1
2
𝑎𝑡2
Solution:
d = vit +
1
2
at2
d = (0m/s) (10s) + (0.5) (8m/s2) (10s)2
d = (0m/s) (10s) + (0.5) (8m/s2) (10s)2
d = (0.5) (8m/s2) (100s2)
d = 400m
Given:
a = 8m/s2
t = 10s
vi = 0m/s
d = ?

20. 20
B. What is the velocity of the car at the end
of 10 seconds?
Given:
a = 8m/s2
t = 10s
vi = 0m/s
vf = ?
Formula:
vf = vi + at

21. 21
Formula:
vf = vi + at
Solution:
vf = vi + at
vf = 0m/s + (8m/s2) (10s)
vf = 80 m/s
Given:
a = 8m/s2
t = 10s
vi = 0m/s
vf = ?

22. Vertical Dimension
or Free Fall
22

23. If you throw an object upward,
what will happen next?
▷ Everything that goes up or thrown
upwards always falls at a constant
acceleration which has a constant
magnitude of 9.8 m/s2
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24. Vertical Dimension
or Free Fall
Problem Solving
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25. 1. Sam is playing a ball on top of a
building but the ball fell and hit the
ground after 2.6 s.
A. What is the final velocity of the
ball just before it hits the ground?
B. How high is the building?
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26. 26
A. What is the final velocity of the ball just
before it hits the ground?
Given:
ag = -9.8 m/s2
vi = 0 m/s
t = 2.6 s
vf = ?
Formula:
vf = vi + agt

27. 27
Solution:
vf = vi + agt
vf = 0m/s + (-9.8 m/s2) (2.6s)
vf = - 26m/s
Given:
ag = -9.8 m/s2
vi = 0 m/s
t = 2.6 s
vf = ?
Formula:
vf = vi + agt

28. 28
B. How high is the building?
Given:
ag = -9.8 m/s2
vi = 0 m/s
t = 2.6 s
vf = ?
Formula:
d = vit +
1
2
agt2

29. 29
Solution:
d = vit +
1
2
agt2
d = (0 m/s) (2.6s) + (-9.8 m/s2) (2.6s) 2
d = (0 m/s) (2.6s) + (-9.8 m/s2) (2.6s) 2
d = (-9.8 m/s2) (6.76s2)
d = - 33m
h = -d = -33m
h = 33m
Given:
ag = -9.8 m/s2
vi = 0 m/s
t = 2.6 s
vf = ?
Formula:
vf = vi + agt

30. 2. Miranda was standing on top of a
building that is 122.5 m high and decided
to drop a rock. What is the final velocity
of the rock just before it hits the ground?
How long will the rock hit the ground?
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31. 31
A. What is the final velocity of the rock just
before it hits the ground?
Given:
d = 122.5 m
vi = 0m/s
a = -9.8 m/s2
vf = ?
Formula:
vf
2 = vi
2 + 2agd

32. 32
Solution:
vf
2 = vi
2 + 2agd
vf
2 = (0m/s)2 + 2(-9.8 m/s2) (122.5 m)
vf
2 = (0m/s)2 + (-19.6 m/s2) (122.5 m)
vf
2 = 2, 401 m2/s2
Formula:
vf
2 = vi
2 + 2agd
Given:
d = 122.5 m
vi = 0m/s
a = -9.8 m/s2
vf = ?

33. 33
B. How long will the rock hit the ground?
Given:
d = 122.5 m
vi = 0m/s
a = -9.8 m/s2
t = ?
Formula:
d = vit +
1
2
agt2

34. 34
Solution:
d = vit +
1
2
agt2
122.5 m = (.5) ( -9.8 m/s2)t2
122.5 m = (- 4.9 m/s2) t2
122.5 𝑚
4.9 m/s2 = t2
25𝑠2 = 𝑡2
t = 5 s
Given:
d = 122.5 m
vi = 0m/s
a = -9.8 m/s2
t = ?
Formula:
d = vit +
1
2
agt2

35. Guide Questions:
1. What is the rate of change of velocity
called?
2. How will you describe uniformly
accelerated motion?
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36. Guide Questions:
3. What is the dimension with respect to the x-
axis plane? In the y-axis plane
4. What are the two directions involved in
Vertical dimension or free fall?
5. What is the constant initial velocity of an
object moving upward? Downward?
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37. TAKE HOME ACTIVITIES
In a clean sheet of paper, write your NAME,
GRADE & SECTION, and TITLE OF MODULE
and answer the following:
Activity 1. Let’s Do It!
Activity 2. I Fill You! Activity
Activity 3. Think and Solve
Activity 4. Write! Right! 37

38. Thanks!
Any questions?
You can find me at:
Facebook: Maden Robie Jane C. Crisostomo
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