SPH 3U Grade 11 U Physics
• This lesson will introduce you to equations of motion called
kinematic equation
• You will be able to use these equations to solve kinematic
related questions
SPH 3U Grade 11 U Physics
What are we going to cover today?
By the end of this lesson,
you will be able to:
Look back at v-t
graphs
The five equations
Sample problems
Homework
SPH 3U Grade 11 U Physics
Recall :
• Slope of a d-t graph is v
• Slope of a
v-t graph is a
• Area under a
• Area under a
v-t graph is Dd
a-t graph is Dv
velocity (m/s [U])
SPH 3U Grade 11 U Physics
Slope = a
vf
vi
area = Dd
ti
tf
time (s)
3s
 What would we need in order to graph the acceleration of a car?
 Motion Detector + computer with specialized software
 slow- motion camera
 Some kind of scale / reference for distance
 Timer
 We would need to create a 𝒅˗𝒕 graph, determine
instantaneous (slope of tangent) and average velocity,
plot the 𝑣 data and make a 𝒗˗𝒕 graph.
 Then we would need to find the slope to determine the
𝒂𝒂𝒗 .
 Or…. If the raw data is provided, we could just
use MATH (algebra) + 𝒐𝒖𝒓 𝑩𝑹𝑨𝑰𝑵
Five Key Uniform Acceleration Equations
 As shown in the previous lessons, graphical
analysis is an important tool physicists to use
to solve problems. However, sometimes we
have enough information to solve problems
algebraically.
 Algebraic methods tend to be quicker and
more convenient than graphical analysis.
 When solving uniform acceleration problems,
choose which equations to use based on the
given and required variables of the problem.
EQUATION 1
 To be able to solve uniform acceleration
problems, we need to derive algebraic
equations from a change in velocity using a
standard 𝒗˗𝒕 graph .
• The 1st Equation stems from
the area under the 𝒗˗𝒕 graph
EQUATION 1
Y=intercept
B = base (run) = ∆t
l = length (run) = ∆t
h = height (rise) = ∆v h = height (rise) = vi
Like
Terms
EQUATION 2
• The 2nd Equation stems from the slope of the
𝒗˗𝒕 graph and rearranging for 𝑣𝑓 .
• What does slope of 𝒗˗𝒕 tell us?
𝒗𝒇 − 𝒗𝒊
𝒂𝒂𝒗 =
• Rearrange for 𝑣𝑓
∆t
X ∆t
X ∆t
𝒂𝒂𝒗 (∆t ) = 𝒗𝒇 − 𝒗𝒊
+ 𝑣𝑖
𝒂𝒂𝒗 (∆t ) + 𝒗𝒊 = 𝒗𝒇
+ 𝑣𝑖
EQUATION 3
 In order to produce the 3rd Equation, we
substitute Equation 2 into Equation 1 for 𝒗𝒇 as
shown below.
EQUATION 3
𝑳𝒊𝒌𝒆 𝑻𝒆𝒓𝒎𝒔
Get rid of the brackets
Via multiplication
=
∆𝒅 =
1
𝟐
(2 𝒗𝒊 + 𝒂𝒂𝒗 ∆t ) ∆t
1
𝒗𝒊 ∆t +
𝒂𝒂𝒗 ∆t 𝟐
𝟐
EQUATION 4 & 5
 Equations 4 and 5 are similarly produced
via substitution
 rearranging Equation 2 for 𝒗𝒊
 rearranging Equation 2 for ∆t.
The 5 Kinematic Equations
 SP #1-3 p.38-39
 It is important to note that these five equations
are used only when an object undergoes
acceleration, a change in velocity.
 If an object is travelling at a constant velocity,
we simply use the equation for constant
velocity from section 1.2.
 REMEMBER: there is only 5 possible variables in any of
these problems.
 Displacement, time, initial velocity, final velocity, &
average acceleration
 Problems in this course can only contain 4 of the
possible 5 variables.
 Each of the 5 equations is therefore missing a different
variable – this ensures that no matter what variable is
missing, we have a formula we can use to solve the
problem.
 This also means we can figure out what formula we need
to use by looking at our givens OR determining which of
the 5 variables listed above we DON’T HAVE
SPH 3U Grade 11 U Physics
Ex 1: A runner jogging west at 5.0 m/s accelerates to
12 m/s while covering 150 m. What is the
acceleration of the runner?
Sol’n:
Given: vi = 5.0 m/s
vf = 12 m/s
Dd = 150 m
Required:a
Analysis: Look for equation with all of these variables.
2010
SPH 3U Grade 11 U Physics
v2f  v2i  2aDd
a
2
2
v

v
 f i
2 Dd
2
2
12  5

2(150)
 .39667
 The acceleration is .40 m/s2 [W]
1.5 Homework
Practice # 1,2 p.39
Questions #1-6 p.39