Uniform Accelerated Motion
Unit 4 : Module 1
REVIEW
•Describe constellations
•Give examples of constellation found
in the Northern and Southern
Hemisphere.
•Why are there some stars that cannot
be seen in some months?
OBJECTIVES
•Describe the Uniformly Accelerated
Motion (UAM) qualitatively and
quantitatively.
•Identify the equations involved in
Uniformly Accelerated Motion in
horizontal dimension
ACTIVITY:
• Following a video presentation about motion
discussed when you were Grade 7 and 8.
• Take down notes to answer the following
questions.
1. What is speed? velocity?
2. What is the difference between speed and
velocity?
3. What is acceleration?
Uniformly Accelerated Motion
Uniformly Accelerated Motion
•constant acceleration
•the object is moving
with constant
acceleration
ESSENTIAL EQUATIONS to KNOW
•Displacement •Velocity
𝑑 = 𝑣𝑡
where:
v = velocity
v = final velocity
v = initial velocity
v = average velocity
v=
𝑑
𝑡
•Average Velocity
𝑣𝑎𝑣𝑒 =
ave
d = displacement
t = time
a = acceleration
2
•Acceleration
f
i
𝑣𝑓 +𝑣𝑖
𝑎=
𝑣𝑓 −𝑣𝑖
𝑡
Uniformly Accelerated Motion
•To find out how
displacement changes
with time when an
object is uniformly
accelerated, we need
to derive its equation
fro the formula of
d, v and a
DERIVATION OF (UAM) EQUATIONS
𝑣𝑓 +𝑣𝑖
• Displacement
2
𝑑 = 𝑣𝑡
𝑣
𝑑 = 𝑣𝑡
•EQUATION 1
• Average Velocity
𝑣𝑎𝑣𝑒 =
𝑣𝑓 +𝑣𝑖
2
𝑑=
𝑣𝑓 +𝑣𝑖
2
t
DERIVATION OF (UAM) EQUATIONS
• Acceleration
𝑎=
𝑣𝑓 −𝑣𝑖
𝑡
𝑣𝑓 𝐢𝐬 𝐨𝐧 𝐭𝐡𝐞 𝐥𝐞𝐟𝐭 𝐬𝐢𝐝𝐞
𝐨𝐟 𝐭𝐡𝐞 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧
𝑣𝑓 − 𝑣𝑖
𝑎=
𝑡
𝑎𝑡 = 𝑣𝑓 − 𝑣𝑖
𝑎𝑡 + 𝑣𝑖 = 𝑣𝑓
𝑣𝑓 = 𝑎𝑡 + 𝑣𝑖
Multiply
Transfer
Change the sign
into +
Rearrange
DERIVATION OF (UAM)
EQUATIONS
• EQUATION 1
𝑑=
𝑣𝑓 +𝑣𝑖
2
t
Final Velocity
𝑣𝑓 = 𝑎𝑡 + 𝑣𝑖
𝑣𝑓 + 𝑣𝑖
𝑑=
t
2
𝑎𝑡 + 𝑣𝑖 + 𝑣𝑖
𝑑=
t
2
2𝑣𝑖 + 𝑎𝑡
𝑑=
t
2
2𝑣𝑖 + 𝑎𝑡
𝑑=
t
2
2𝑣𝑖 𝑡 + 𝑎𝑡 2
𝑑=
2
2𝑣𝑖 𝑡 𝑎𝑡 2
𝑑=
+
2
2
𝑎𝑡
𝑑 = 𝑣𝑖 𝑡 +
2
2
DERIVATION OF (UAM) EQUATIONS
•Acceleration
𝑎=
𝑣𝑓 −𝑣𝑖
𝑡
•Acceleration
𝑎=
𝑣𝑓 −𝑣𝑖
𝑡
Time
𝑡=
𝑣𝑓 −𝑣𝑖
𝑎
DERIVATION OF (UAM)
EQUATIONS
• Displacement
𝑑 = 𝑣𝑡
𝑑 = 𝑣𝑡
𝑣𝑓 + 𝑣𝑖 𝑣𝑓 − 𝑣𝑖
𝑑=
2
𝑎
𝑣𝑓 2 − 𝑣𝑖 2
𝑑=
2𝑎
2𝑎𝑑 = 𝑣𝑓 2 − 𝑣𝑖 2
• Average Velocity
𝑣𝑎𝑣𝑒 =
𝑣𝑓 +𝑣𝑖
2
Time
𝑡=
2𝑎𝑑 + 𝑣𝑖 2 = 𝑣𝑓 2
Rearranging
𝑣𝑓 −𝑣𝑖
𝑎
2
𝑣𝑓 = 2𝑎𝑑 + 𝑣𝑖
2
Uniformly Accelerated Motion (UAM)
Equations
𝑑=
𝑣𝑓 +𝑣𝑖
2
𝑎𝑡
𝑑 = 𝑣𝑖 𝑡 +
2
t
2
𝑣𝑓 = 2𝑎𝑑 + 𝑣𝑖
2
2
PROBLEM-SOLVING ON UAM
• To apply these derived equations, study the following problems.
An airplane
accelerates from rest
on a runway at 5.50
𝟐
m/𝒔 for 20.25
seconds, until it
finally takes off the
ground. What is the
distance covered
before takeoff?
GIVEN:
a = 5.50 m/𝒔
t = 20.25 s
𝒗𝒊 = 𝟎 (at rest)
𝟐
FIND:
d=?
USE:
since it
satisfies all
the given.
𝑎𝑡 2
𝑑 = 𝑣𝑖 𝑡 +
2
An airplane accelerates from rest on a runway at 5.50 m/s 2
for 20.25 seconds, until it finally takes off the ground. What
is the distance covered before takeoff?
GIVEN:
a = 5.50 m/𝒔𝟐
t = 20.25 s
𝒗𝒊 = 𝟎 (at rest)
FIND:
d=?
𝑎𝑡 2
𝑑 = 𝑣𝑖 𝑡 +
2
𝑚
(5.50 2 )(20.25𝑠)2
𝑠
𝑑 = (0𝑚/𝑠)(20.25𝑠) +
2
𝑚
2
5.50
410.0625𝑠
𝑚
2
𝑠
𝑑= 0 2 +
𝑠
2
2, 255.34375 𝑚
𝑑=
2
𝒅 = 𝟏, 𝟏𝟐𝟕. 𝟔𝟕 𝒎
PROBLEM-SOLVING ON UAM
• To apply these derived equations, study the following problems.
From rest, a jeepney
accelerates
uniformly over a
time of 3.25 seconds
and covers a
distance of 15 m.
Determine the
acceleration of the
jeepney.
GIVEN:
t = 3.25 s
d = 15 m
𝒗𝒊 = 𝟎 (at rest)
FIND:
a=?
USE:
since it
satisfies all
the given.
𝑎𝑡 2
𝑑 = 𝑣𝑖 𝑡 +
2
From rest, a jeepney accelerates uniformly over GIVEN:
a time of 3.25 seconds and covers a distance of
t = 3.25 s
15 m. Determine the acceleration of the
d = 15 m
jeepney.
𝒗𝒊 = 𝟎 (at rest)
𝑎𝑡 2
𝑑 = 𝑣𝑖 𝑡 +
2
𝑎 (3.25𝑠)2
15𝑚 = (0𝑚/𝑠)(3.25𝑠) +
2
𝑚
𝑎
15𝑚 = 0 2 +
𝑠
15𝑚 =
𝑎
30 = 𝑎
10.5625𝑠 2
2
10.5625𝑠 2
2
10.5625𝑠 2
FIND:
a=?
30 𝑚
𝑎 10.5625𝑠 2
=
2
10.5625𝑠
10.5625𝑠 2
𝒂 = 𝟐. 𝟖𝟒 𝒎/𝒔𝟐
ACTIVITY:
SOLVE the following problems using the equation of
Uniformly Accelerated Motion.
1. An airplane accelerates down a runway at 3.20
m/s2 for 32.8 s until is finally lifts off the ground.
Determine the distance traveled before takeoff.
2. A car starts from rest and accelerates uniformly
over a time of 5.21 seconds for a distance of 110
m. Determine the acceleration of the car.
ACTIVITY:
1. An airplane
accelerates
down a
runway at
3.20 m/s2 for
32.8 s until is
finally lifts off
the ground.
Determine the
distance
traveled
before takeoff.
SOLVE the following problems using the
equation of Uniformly Accelerated Motion.
𝑎𝑡 2
𝑑 = 𝑣𝑖 𝑡 +
2
𝑚
(3.20 2 )(32.8𝑠)2
𝑠
𝑑 = (0𝑚/𝑠)(32.8𝑠) +
2
𝑚
2
3.20
1,
075.84𝑠
𝑚
2
𝑠
𝑑= 0 2 +
𝑠
2
GIVEN:
a= 3.20 m/s2
3, 442.688 𝑚
𝑑=
t = 32.8 s
2
𝒗𝒊 = 𝟎 (at rest)
𝒅 = 𝟏, 𝟕𝟐𝟏. 𝟑𝟒𝟒𝒎
FIND:
d=?
𝑎𝑡 2
𝑑 = 𝑣𝑖 𝑡 +
2
ACTIVITY:
2. A car starts
from rest and
accelerates
uniformly over
a time of 5.21
seconds for a
distance of 110
m. Determine
the acceleration
of the car.
GIVEN:
t =5.21s
d = 110 m
𝒗𝒊 = 𝟎 (at rest)
𝑎 (5.21𝑠)2
110𝑚 = (0𝑚/𝑠)(5.21𝑠) +
2
𝑚
𝑎
110𝑚 = 0 2 +
𝑠
FIND:
a=?
27.1441𝑠 2
2
𝑎
27.1441𝑠 2
2
220 𝑚 = 𝑎
27.1441𝑠 2
110 𝑚 =
220 𝑚
𝑎 27.1441𝑠 2
=
2
27.1441𝑠
27.1441𝑠 2
𝒂 = 𝟖. 𝟏𝟎 𝒎/𝒔𝟐
• A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47
seconds. Determine the acceleration of the car and the distance
traveled.
• A bike accelerates uniformly from rest to a speed of 7.10 m/s over
ad
• A plane has a takeoff speed of 88.3 m/s and requires 1365 m to
reach that speed. Determine the acceleration of the plane and the
time required to reach this speed.istance of 35.4 m. Determine the
acceleration of the bike.