Quarter Exam Revision
2022/2023
Dr. Marwan Heaba
01025900678
“Success is not final, failure is not fatal: it is the courage to continue that counts”
Winston Churchill
Grade 12 Physics
Dr/Marwan Heaba
Physics

Any problem that deals with temperature, size, motion, position, shape, or color involves physics.

Physics principles can be used to make predictions about a broad range of phenomena.
For example, the same physics principles that are used to describe the interaction between two planets
can be used to describe the motion of a satellite orbiting Earth.

Many of the inventions, appliances, tools, and building we live with today are made possible by
the application of physics principles.
Branches of physics
Name
Subjects
Example
Mechanics
Motion, its causes, and interactions
between objects.
Friction, falling objects, weight, and
spinning object.
Thermodynamics
Heat and temperature.
Melting and freezing process,
refrigerators, and engines.
Vibrations and wave
phenomena
Specific types of repetitive motions.
Sound, pendulums, and springs.
Optics
Light.
Lenses, mirrors, and colors.
Electromagnetism
Electricity and magnetism.
Electrical charge and permanent
magnets.
Relativity
Particles moving at any speed
including very high speeds.
Particle collisions, particles
accelerators, and nuclear energy.
Quantum mechanics
Behavior of submicroscopic
particles.
Atom and its parts.
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Scientific Method
Models
 A pattern, representation, or description designed to show the structure or workings of an object, or system.
 Physicists often use models to explain the most fundamental features of various phenomena.
System
 A set of interacting components considered to be a distinct physical entity for the purpose of study.
Hypothesis
 Explanation of observations and that can be tested.
 A hypothesis must be tested in a controlled experiment.
Controlled experiment
 Testing only one factor at a time by using a comparison of a control group with an experimental group.
Galileo’s Hypothesis
 Behavior of falling objects in order to develop a hypothesis about how objects fall.
 In the absence of air resistance, all objects fall at the same rate regardless of their mass.
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Grade 12 Physics
Dr/Marwan Heaba
Measurement


Physicists perform experiments to test hypotheses about how changing one variable in a situation
affects another variable.
An accurate analysis of such experiments requires numerical measurements.
Dimension


The description of what kind of physical quantity is represented by a certain measurement.
Ex: length, mass, and time
Measuring unit
 The description of how much of a physical quantity is represented by a certain numerical measurement.
 Ex: meter, gram, and second
International System of units (SI)



In SI, the standard measurement system for science, there are seven base units.
Each base unit describes a single dimension, such as length, mass, or time.
Derived units are formed by combining base units with multiplication or division. For example, speeds
are typically expressed in units of meters per second (m/s).
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Physical Quantity
Name of Unit
Symbol
(Abbreviation)
Length
Meter
M
Mass
Kilogram
Kg
Time
Second
S
Temperature
Kelvin
K
Amount of substance
Mole
mol
Electric current
Ampere
A
Intensity of light
candela
cd
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Grade 12 Physics
Dr/Marwan Heaba
Dimensions and Units

Measurements used in calculations should also have the same units.

When determining area by multiplying length and width,
be sure the measurements are expressed in the same units.
 The distance between sun and earth is 150, 000 000 000 meters
 The distance between Alex and Cairo is 220,000 meters
 The length of human is about 1.6 – 1.8 meters
 The length of insect is about 0.01 meters
 The radius of atom is about 0.000 000 000 000 03 meters
SI Prefixes

In SI, units are combined with prefixes that symbolize certain powers of 10
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Grade 12 Physics
Dr/Marwan Heaba
Mathematics and Physics

Tables, graphs, and equations can make data easier to understand.
Galileo’s hypothesis all objects fall at the same rate in the absence of air resistance
Change in position in meters = 4.9  (Time in seconds) 2
∆
X = 4.9 (∆t) 2

Greek letter ∆ (delta) is often used to mean “difference or change in,”

Greek letter Σ (sigma) is used to mean “sum” or “total.”
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Grade 12 Physics

Dr/Marwan Heaba
Which of the following statements is true of any valid physical equation?
a. Both sides have the same dimensions.
b. Both sides have the same variables.
c. There are variables but no numbers.
d. There are numbers but no variables
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One Dimensional Motion

Kind of motion that take place in one direction.

Ex: Motion of train on a straight track.

To measure motion, you must choose a frame of reference.

A frame of reference is a system for specifying the precise location of objects in space and time.
Displacement

Length of the straight line drawn from its initial position to the object’s final position.

Displacement is not always equal to the distance traveled.

SI unit of displacement is the meter, m.

Vector quantity.
Distance

It is the actual path moved in it.

SI unit of distance is the meter, m.

Scalar quantity.
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Velocity

Displacement per unit time.

SI unit of velocity is the meters per second, m/s

(From Km/hr to m/s multiply by 𝟔𝟎 𝒙 𝟔𝟎 )
𝟏𝟎𝟎𝟎
V=

 (𝐗) 𝐝𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭
 (𝐭) 𝐭𝐢𝐦𝐞
Average velocity equal total displacement divided by the time interval
Vavg =
Total displacement
Total time
Speed

Distance covered in unit time.

SI unit of speed is the meters per second, m/s
S=
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 (𝐗) 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞
 (𝐭) 𝐭𝐢𝐦𝐞
01025900678
Grade 12 Physics
Dr/Marwan Heaba
Consider a car trip to a friend’s house 370 km to the west along a straight highway.
If you left your house at 10 A.M. and arrived at your friend’s house at 3 P.M.,
Your average velocity would be as follows:
Vavg =
Vavg =
−𝟑𝟕𝟎 𝐤𝐦
𝟓 𝐡𝐫
𝐓𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭
𝐓𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞
= - 74 Km/hr = 74 Km/hr west
During a race on level ground, Andra runs with an average velocity of 6.02 m/s to the east.
What is Andra’s displacement after 137 s?
Vavg =
𝐓𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭
𝐓𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞
∆x = Vavg ∆t
∆x = (6.02 m/s) (137 s) = 825 m to east
Heather and Matthew walk with an average velocity of 0.98 m/s eastward.
If it takes 34 min to walk to the store, what is their displacement?
Vavg =
𝐓𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭
𝐓𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞
∆x = Vavg ∆t
∆t = 34 * 60 = 2040 sec
∆x = (0.98 m/s) (2040 s) = 2000 m to east
It takes you 9.5 min to walk with an average velocity of 1.2 m/s to the north from the bus stop to
the museum entrance. What is your displacement?
Vavg =
𝐓𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭
𝐓𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞
∆x = Vavg ∆t
∆t = 9.5 * 60 = 570 sec
∆x = (1.2 m/s) (570 s) = 684 m to north
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Grade 12 Physics
Dr/Marwan Heaba
If Joe rides his bicycle in a straight line for 15 min with an average velocity of 12.5 km/h south,
How far has he ridde?
Vavg =
𝐓𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭
𝐓𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞
∆x = Vavg ∆t
∆t = 15 / 60 = 0.25 hr
∆x = (12.5 km/hr) (0.25 hr) = - 3.125 km = 3.125 km to south
Simpson drives his car with an average velocity of 48.0 km/h to the east.
How long will it take him to drive 144 km on a straight highway?
Vavg =
𝐓𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭
𝐓𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞
∆t = ∆x / Vavg
∆t = (144 km) / (48 km/hr) = 3 hr
How much time would Simpson save by increasing his average velocity to 56.0 km/h to the east?
∆t = ∆x / Vavg
∆t = (144 km) / (56 km/hr) = 2.57 hr
He would save 3 hr - 2.57 hr = 0.43 hr = 25.8 min
A bus travels 280 km south along a straight path with an average velocity of 88 km/h to the south. The
bus stops for 24 min. Then, it travels 210 km south with an average velocity of 75 km/h to the south.
a. How long does the total trip last?
b. What is the average velocity for the total trip?
Vavg =
𝐓𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭
𝐓𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞
∆t = ∆x / Vavg
∆t = (280 km) / (88 km/hr) = 3.18 hr
∆t = ∆x / Vavg
∆t = (210 km) / (75 km/hr) = 2.8 hr
∆t = 3.18 + 2.8 + (24 min / 60) = 6.38 hr
𝟐𝟖𝟎 + 𝟐𝟏𝟎
Vavg =
= -76.8 km/hr = 76.8 km/hr to south
𝟔.𝟑𝟖
Car A travels from New York to Miami at a speed of 25 m/s. Car B travels from New York to Chicago, also
at a speed of 25 m/s. Are the velocities of the cars equal? Explain
Not Equal. As the direction from New York to Miami (due South) is different from the direction from
New York to Chicago (due West), the 2 velocities are not equal
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Grade 12 Physics
Dr/Marwan Heaba
What is the shortest possible time in which a bacterium could travel a distance of 8.4 cm across a Petri
dish at a constant speed of 3.5 mm/s?
S=
𝐓𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞
𝐓𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞
∆x = 8.4 cm ---------> 8.4 x 10 = 84 mm
∆t = ∆x / S
∆t = (84 mm) / (3.5 mm/s) = 24 sec
A child is pushing a shopping cart at a speed of 1.5 m/s.
How long will it take this child to push the cart down an aisle with a length of 9.3 m?
S=
𝐓𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞
𝐓𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞
∆t = ∆x / S
∆t = (9.3 m) / (1.5 m/s) = 6.2 sec
An athlete swims from the north end to the south end of a 50.0 m pool in 20.0 s and makes the return trip
to the starting position in 22.0 s.
a. What is the average velocity for the first half of the swim?
b. What is the average velocity for the second half of the swim?
c. What is the average velocity for the roundtrip?
a)
Vavg =
b)
𝐓𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭
c)
Vavg =
𝐓𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞
Vavg = (50 m) / (20 sec) = 2.5 m/s south
𝐓𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭
𝐓𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞
Vavg = (50 m) / (22 sec) = 2.27 m/s north
Vavg =
𝐓𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭
𝐓𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞
Vavg = (100 m) / (42 sec) = 2.38 m/s
Two students walk in the same direction along a straight path,
at a constant speed—one at 0.90 m/s and the other at 1.90 m/s.
a. Assuming that they start at the same point and the same time, how much sooner does the faster student
arrive at a destination 780 m away?
b. How far would the students have to walk so that the faster student arrives 5.50 min before the slower
student?
Vavg =
𝐓𝐨𝐭𝐚𝐥 𝐝𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭
𝐓𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞
∆t = ∆x / Vavg
∆t = (780 m) / (1.9 m/sec) = 410.5 sec = 6.8 min
∆x faster = ∆x slower = Vavg x ∆t
0.9(t+5.5) = 1.9t
0.9t + 4.95 = 1.9t
4.95 = 1.9t-0.9t
4.95 = t
Faster should walk 1.9 X t in seconds =1.9 (4.95 X 60) = 564.3 meters
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Acceleration

The rate at which velocity changes over time.

An object accelerates if its speed, direction, or both change.

Acceleration has direction and magnitude. Thus, acceleration is a vector quantity.

Measuring unit of acceleration = m/sec2  m.sec-2 or Km/hr2  Km.hr-2
𝐂𝐡𝐚𝐧𝐠𝐞 𝐢𝐧 𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲
Average acceleration = 𝐓𝐢𝐦𝐞 𝐫𝐞𝐪𝐮𝐢𝐫𝐞𝐝 𝐟𝐨𝐫 𝐜𝐡𝐚𝐧𝐠𝐞
aavg =

𝐕
𝐓
=
𝐕𝐟 − 𝐕𝐢
𝐓𝐟 − 𝐓𝐢
Acceleration could be positive or negative.
☺
If the velocity increases, the acceleration is positive.
☺
If the velocity decreases, the acceleration is negative.
The negative acceleration is called deceleration or retardation.
 When the velocity in the positive direction is increasing, the acceleration is positive, as at A.
 When the velocity is constant, there is no acceleration, as at B.
 When the velocity in the positive direction is decreasing, the acceleration is negative, as at C
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Dr/Marwan Heaba
Problem
A shuttle bus slows down with an average acceleration of 1.8 m/s2
How long does it take the bus to slow from 9.0 m/s to a complete stop?
Given
Vi = 9.0 m/s
Unknown
∆t = ?
Vf = 0 m/s aavg = −1.8 m/s2
aavg =
t =
Solution
𝐕
𝐚
=
𝐕𝐟 − 𝐕𝐢
𝐚
𝐕
𝐭
=
𝟎 𝐦/𝐬 − 𝟗 𝐦/𝐬
− 𝟏.𝟖 𝐦/𝒔𝟐
t = 5 sec
Problem
Marissa’s car accelerates uniformly at a rate of +2.60 m/s2. How long does it take Marissa’s
car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s?
Given
Vi = 24.6 m/s Vi = 26.8 m/s
Unknown
∆t = ?
aavg = 2.6 m/s2
aavg =
Solution
∆t =
𝐕𝐟 − 𝐕𝐢
𝐚
=
𝐕𝐟 − 𝐕𝐢
𝐓
𝟐𝟔.𝟖 − 𝟐𝟒.𝟔
𝟐.𝟔
= 0.85 s
t = 0.85 sec
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Dr/Marwan Heaba
Problem
As the shuttle bus comes to a sudden stop to avoid hitting a dog,
It accelerates uniformly at − 4.1 m/s2 as it slows from 9.0 m/s to 0.0 m/s.
Find the time interval of acceleration for the bus.
Given
Vi = 9.0 m/s Vf = 0 m/s aavg = −4.1 m/s2
Unknown
∆T = ?
aavg =
t =
Solution
𝐕
𝐚
=
𝐕𝐟 − 𝐕𝐢
𝐚
𝐕
𝐓
=
𝟎 𝐦/𝐬 − 𝟗 𝐦/𝐬
− 𝟒.𝟏 𝐦/𝐬𝟐
t = 2.19 sec
Problem
A car traveling at 7.0 m/s accelerates uniformly at 2.5 m/s2 to reach a speed of 12.0 m/s.
How long does it take this acceleration to occur?
Given
Vi = 7.0 m/s
Unknown
∆T = ?
Vf = 12.0 m/s
aavg = 2.5 m/s2
aavg =
Solution
t =
𝐕
𝐚
=
𝐕𝐟 − 𝐕𝐢
𝐚
=
𝐕
𝐓
𝟏𝟐 𝐦/𝐬 − 𝟕 𝐦/𝐬
𝟐.𝟓 𝐦/𝒔𝟐
t = 2 sec
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Problem
With an average acceleration of −1.2 m/s2, How long will it take a cyclist to bring a bicycle
with an initial speed of 6.5 m/s to a complete stop?
Given
Vi = 6.5 m/s Vf = 0 m/s
Unknown
∆T = ?
aavg = −1.2 m/s2
aavg =
t =
Solution
𝐕
𝐚
=
𝐕𝐟 − 𝐕𝐢
𝐚
=
𝐕
𝐓
𝟎 𝐦/𝐬 − 𝟔.𝟓 𝐦/𝐬
− 𝟏.𝟐 𝐦/𝒔𝟐
t = 5.416 sec
Problem
Turner’s treadmill runs with a velocity of −1.2 m/s and speeds up at regular intervals during
a half-hour workout. After 25 min, the treadmill has a velocity of − 6.5 m/s. What is the
average acceleration of the treadmill during this period?
Given
Vi = - 1.2 m/s
Unknown
aavg = ?
Vf = - 6.5 m/s
∆t = 25 min  25 * 60 = 1500 sec
aavg =
Solution
aavg =
𝐕𝐟 − 𝐕𝐢
𝐓
=
𝐕
𝐓
(− 𝟔.𝟓 𝐦/𝐬) − (−𝟏.𝟐 𝐦/𝐬)
𝟏𝟓𝟎𝟎
aavg = - 0.00353 m/s2
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Dr/Marwan Heaba
Problem
A plane starting at rest at one end of a runway undergoes a uniform acceleration of 4.8
m/s2 for 15 s before takeoff. What is its speed at takeoff?
How long must the runway be for the plane to be able to take off?
Given
Vi = 0 m/s ∆t = 15 sec
Unknown
Vf = ? ∆X = ?
aavg = 4.8 m/s2
Vf = Vi + at
Vf = (0) + (4.8) (15)
Vf = 72 m/s
Solution
𝟏
X = (Vi + Vf) t
𝟐
Solution
𝟏
X= (0 + 72 ) (15) = 540 m
𝟐
Problem
Suppose a treadmill has an average acceleration of 4.7 × 10−3 m/s2
a. How much does its speed change after 5.0 min?
b. If the treadmill’s initial speed is 1.7 m/s, what will its final speed be?
Given
Vi = 0 m/s aavg = 4.7 x 10-3
Solution
Solution
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∆t = 5 min  5 * 60 = 300 sec
Vf = Vi + at
Vf = 4.7 x 10-3 x 300 = 1.41 m/s
Vf = Vi + at
Vf = 1.7 + (4.7 x 10-3 x 300) = 3.11 m/s
01025900678
Grade 12 Physics
Dr/Marwan Heaba
𝟏
𝟏
X=Vi t + a(t)2
X= (Vi + Vf) t
𝟐
𝟐
Problem
A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using
its parachute and braking system, and comes to rest 5.5 s later.
Find the distance that the car travels during braking.
Given
Vi = 42 m/s Vf = 0 m/s ∆t = 5.5 s
Unknown
∆X = ?
𝟏
X= (Vi + Vf) (t)
𝟐
𝟏
X= (42 m/s + 0 m/s) (5.5 s)
Solution
𝟐
 X = 115.5 m
Problem
A car accelerates uniformly from rest to a speed of 6.6 m/s in 6.5 s.
Find the distance the car travels during this time.
Given
Vi = 0 m/s Vf = 6.6 m/s ∆t = 6.5 s
Unknown
∆X = ?
𝟏
X= (Vi + Vf) (t)
𝟐
𝟏
Solution
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X= (0 m/s + 6.6 m/s ) (6.5 s)
𝟐
 X = 21.45 m
01025900678
Grade 12 Physics
Dr/Marwan Heaba
Problem
When Maggie applies the brakes of her car, the car slows uniformly from
15.0 m/s to 0.0 m/s in 2.50 s. How many meters before a stop sign must she apply her
brakes in order to stop at the sign?
Given
Vi = 15 m/s
Vf = 0 m/s
∆t = 2.5 s
Unknown
∆X = ?
𝟏
X= (Vi + Vf) (t)
𝟐
𝟏
Solution
X= (15 m/s + 0 m/s ) (2.5 s)
𝟐
 X = 18.75 m
Problem
A driver in a car traveling at a speed of 21.8 m/s sees a cat 101 m away on the road. How long
will it take the car to accelerate uniformly to a stop in exactly 99 m?
Given
Vi = 21.8 m/s
Vf = 0 m/s
∆X = 99 m
Unknown
∆t = ?
1
X= (Vi + Vf) (t)
Solution
20 | Page
t= 1
2
2
X
(Vf + Vi )
=1
2
99 m
(0+ 21.8 m/s )
 t = 9.08 s
01025900678
Grade 12 Physics
Dr/Marwan Heaba
Problem
A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for
3.2 km in 3.5 min. How fast (in m/s) is the car moving after this time?
Given
Vi = 6.4 m/s
∆t = 3.5 min  3.5 x 60 = 210 sec
∆X = 3.2 km  3.2 x 1000 = 3200 m
Unknown
Vf = ?
1
X= (Vi + Vf) (t)
(Vi + Vf) =
Solution
2X
t
2
=
2 x 3200 m
210 s
= 30.47 m/s
Vf = 30.47 – 6.4 = 24.07 m/s
Problem
A car with an initial speed of 6.5 m/s accelerates at a uniform rate of 0.92 m/s2 for
3.6 s. Find the final speed and the displacement of the car during this time.
Given
Vi = 6.5 m/s
∆t = 3.6 sec
aavg = 0.92 m/s2
Unknown
Vf = ?
Vf = Vi + at
Solution
21 | Page
Vf = 6.5 m/s + (0.92 m/s2 x 3.6 s) = 9.81 m/s
01025900678
Grade 12 Physics
Dr/Marwan Heaba
Problem
An automobile with an initial speed of 4.30 m/s accelerates uniformly at the rate of 3.00
m/s2. Find the final speed and the displacement after 5.00 s.
Given
Vi = 4.3 m/s
∆t = 5 sec
aavg = 3 m/s2
Unknown
Vf = ?
Vf = Vi + at
Solution
Vf = 4.3 m/s + (3 m/s2 x 5 s) = 19.3 m/s
Problem
A car starts from rest and travels for 5.0 s with a constant acceleration of 1.5 m/s 2
What is the final velocity of the car? How far does the car travel in this time interval?
Given
Vi = 0 m/s
∆t = 5 sec
aavg = 1.5 m/s2
Unknown
Vf = ?
Solution
Vf = Vi + at
Vf = 0 m/s + (1.5 m/s2 x 5 s) = 7.5 m/s
1
X=Vi t + a(t)2
2
1
X= (1.5) (5)2 = 18.75 m
2
22 | Page
01025900678
Grade 12 Physics
Dr/Marwan Heaba
Vf2 = Vi2 + 2ax
Problem
A person pushing a stroller starts from rest, uniformly accelerating at a rate of
0.5 m/s2. What is the velocity of the stroller after it has traveled 4.75 m?
Given
Vi = 0 m/s
a = 0.5 m/s2
∆X = 4.75 m
Unknown
Vf = ?
Vf2 = Vi2 + 2ax
Vf2 = 0 + 2 (0.5 m/s2) (4.75 m)
Solution
Vf = √4.75 = 2.18 m/s
Problem
A person pushing a stroller starts from rest, uniformly accelerating at a rate of
0.5 m/s2. What is the velocity of the stroller after it has traveled 6.32 m?
Given
Vi = 0 m/s
a = 0.5 m/s2
∆X = 6.32 m
Unknown
Vf = ?
Solution
Vf2 = Vi2 + 2ax
Vf2 = 0 + 2 (0.5 m/s2) (6.32 m)
Vf = √4.75 = 2.51 m/s
23 | Page
01025900678
Grade 12 Physics
Dr/Marwan Heaba
Problem
A car accelerates uniformly in a straight line from rest at the rate of 2.3 m/s 2.
a. What is the speed of the car after it has traveled 55 m?
b. How long does it take the car to travel 55 m?
Given
Vi = 0 m/s
a = 2.3 m/s2
∆X = 55 m
Unknown
Vf = ?
Vf2 = Vi2 + 2ax
Vf2 = 0 + 2 (2.3 m/s2) (55 m)
Vf = √253 = 15.9 m/s
Solution
t =
V
aavg
=
Vf − Vi
aavg
=
15.9 m/s − 0 m/s
2.3 m/𝑠 2
t = 6.91 sec
Problem
A motorboat accelerates uniformly from a velocity of 6.5 m/s to the west to a velocity of 1.5
m/s to the west. If its acceleration was 2.7 m/s2 to the east,
How far did it travel during the acceleration?
Given
Vi = 6.5 m/s west
Vf = 1.5 m/s west
a = 2.7 m/s2 west
Unknown
∆X = ?
Solution
24 | Page
Vf2 = Vi2 + 2ax
(1.5)2 = (6.5)2 + 2 (2.7) (x)
x = 7.4 m west
01025900678
Grade 12 Physics
Dr/Marwan Heaba
Problem
An aircraft has a liftoff speed of 33 m/s. What minimum constant acceleration does this
require if the aircraft is to be airborne after a take-off run of 240 m?
Given
Vi = 0 m/s
Vf = 33 m/s
∆X = 240 m
Unknown
a=?
Solution
Vf2 = Vi2 + 2ax
(33)2 = (0)2 + 2 a (240)
a = 2.27 m/s2
Problem
A certain car is capable of accelerating at a uniform rate of 0.85 m/s2.
What is the magnitude of the car’s displacement as it accelerates uniformly from a speed of
83 km/h to one of 94 km/h?
Vi = 83 km/h  83 x
Given
Vf = 94 km/h  94 x
aavg = 0.85 m/s2
Unknown
∆X = ?
Solution
25 | Page
1000
60 x 60
1000
60 x 60
= 23.055
= 26.11
Vf2 = Vi2 + 2ax
(26.11 m/s)2 = (23.05 m/s)2 + 2 (0.85 m/s2) (x)
x = 88 m
01025900678
Grade 12 Physics
Dr/Marwan Heaba
Problem
A car traveling initially at +7.0 m/s accelerates uniformly at the rate of +0.80 m/s 2 for a
distance of 245 m. What is its velocity at the end of the acceleration?
Given
Vi = 7 m/s
∆X = 245 m
aavg = 8 m/s2
Unknown
Vf = ?
Vf2 = Vi2 + 2ax
Vf2 = (7 m/s)2 + 2 (8) (245)
Vf = 63 m/s
Solution
a=
Vf − Vi
T
Vf = Vi + at
Vf2 = Vi2 + 2ax
1
X=Vi t + a(t)2
2
1
X= (Vi + Vf) (t)
2
26 | Page
01025900678
Grade 12 Physics
Dr/Marwan Heaba
Free fall









The motion of a body when gravity is acting on the body.
The acceleration on an object in free fall is called the acceleration due to gravity, or free fall.
Free fall acceleration is denoted with the symbols ag (generally) or g (on Earth’s surface).
Free fall acceleration is the same for all objects, regardless of mass.
Free-fall acceleration on Earth’s surface is –9.81 m/s2 at all points in the object’s motion.
Consider a ball thrown up into the air.
Moving upward: velocity is decreasing, acceleration is –9.81 m/s2
Top of path: velocity is zero, acceleration is –9.81 m/s2
Moving downward: velocity is increasing, acceleration is –9.81 m/s2
Problem
A robot probe drops a camera off the rim of a 239 m high cliff on Mars,
where the free-fall acceleration is −3.7 m/s2
a. Find the velocity with which the camera hits the ground.
b. Find the time required for it to hit the ground.
Given
Vi = 0 m/s
∆X = 239 m
ag = - 3.7 m/s2
Unknown
Vf = ? t = ?
Vf2 = Vi2 + 2ax
Vf2 = 2 (-3.7) (-239)
Vf = - 42.05 m/s
Solution
t =
27 | Page
Vf − Vi
ag
=
−42.05 m/s − 0 m/s
−3.7 m/𝑠 2
= 11.36 s
01025900678
Grade 12 Physics
Dr/Marwan Heaba
Problem
A flowerpot falls from a windowsill 25.0 m above the sidewalk.
a. How fast is the flowerpot moving when it strikes the ground?
b. How much time does a passerby on the sidewalk below have to move out of the way
before the flowerpot hits the ground?
Given
Vi = 0 m/s
∆X = - 25 m
ag = - 9.81 m/s2
Unknown
Vf = ? t = ?
Vf2 = Vi2 + 2ax
Vf2 = 2 (-9.81) (-25)
Vf = - 22.15 m/s
Solution
t =
Vf − Vi
ag
=
−22.15 m/s − 0 m/s
−9.81 m/𝑠 2
= 2.26 s
Problem
A tennis ball is thrown vertically upward with an initial velocity of +0.80 m/s.
a. What will the ball's speed be when it returns to its starting point?
b. How long will the ball take to reach its starting point?
Given
Vi = 0 m/s
Unknown
Vf = ? t = ?
Solution
∆X = - 0.8 m
ag = - 9.81 m/s2
Vf = - 8 m/s
the same as the equal but opposite velocity as the starting one
t =
Vf − Vi
ag
=
0 m/s − 8 m/s
−9.81 m/𝑠 2
= 0.82 s
Time = 0.82 x 2 = 1.62 Don't forget to double it!
28 | Page
01025900678