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UNIFORMLY ACCELERATED
MOTION
PREPARED BY:
MADEN ROBIE JANE C. CRISOSTOMO
INTERN
Activity 1.
Word Association
Direction: Think of word/s or situations that can be
associated from the words below.
2
DISTANCE
▷ the actual length
covered by an object
▷ It is a scalar quantity
meaning it only has a
numerical value
3
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
SPEED
▷ Speed is how fast you
are traveling
▷ eg. riding a car with a
speed of 20 m/s
5
VELOCITY
▷ the speed in a given
direction.
▷ eg. riding a car with a
speed of 20 m/s going
North
6
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
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
Uniformly Accelerated
Motion
▷ is motion of an object where the
acceleration is constant.
9
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
Countdown to 5
5 variables
4 formulas
3 able to find 3 variables
2 you can obtain 2 remaining variables
1 happy physics student
11
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
4 Formulas
13
1. 𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡
2. 𝑑 = 𝑡
𝑣𝑓− 𝑣𝑖
2
3. 𝑑 = 𝑣𝑖𝑡 +
1
2
𝑎𝑡2
4. vf2 = vi2 + 2ad
Horizontal Dimension
Problem Solving
14
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?
15
Given:
a = 5.50 𝑚/𝑠2
t = 20.25 s
vi = 0 m/s
d =?
Formula:
𝑑 = 𝑣𝑖𝑡 +
1
2
𝑎𝑡2
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
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?
17
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
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
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
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 = ?
Vertical Dimension
or Free Fall
22
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
23
Vertical Dimension
or Free Fall
Problem Solving
24
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?
25
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
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
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
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
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?
30
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
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
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
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
Guide Questions:
1. What is the rate of change of velocity
called?
2. How will you describe uniformly
accelerated motion?
35
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?
36
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
Thanks!
Any questions?
You can find me at:
Facebook: Maden Robie Jane C. Crisostomo
38

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Grade 9 Q4-W1_Uniformly Accelerated Motion.pptx

  • 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 6
  • 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 11
  • 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
  • 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? 15 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? 17
  • 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 = ?
  • 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 23
  • 24. Vertical Dimension or Free Fall Problem Solving 24
  • 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? 25
  • 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? 30
  • 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? 35
  • 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? 36
  • 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 38

Editor's Notes

  1. What can you say about the readings of the speedometer after every one minute? What happens after one-minute passes by based on the table?