KINEMATICS – Motion
with constant
acceleration
Level 1 Physics
Objectives and Essential
Questions
 Objectives
 Define and apply definitions
of displacement, avg.
velocity, instantaneous
velocity, and average
acceleration
 Demonstrate proficiency in
problem solving using
kinematic equations
including problems involving
free-fall
 Analyze motion of graphs
qualitatively and
quantitatively.
 Essential Questions
 What is displacement? How
does it differ from distance?
 How is displacement affected
by time, velocity, and
acceleration?
 How is scientific data
displayed?
SYMBOLS
x, y
Displacement
t
Time
vo
Initial velocity
v
Finial velocity
a
Acceleration
g
Acceleration due to gravity
Equation #1
v v  v o
a

 v  v o  at
t
t
v  vo  at


EXAMPLE #1
A speedboat has a constant acceleration of +2.0 m/s2. If the initial
velocity of the boat is 6.0 m/s, how fast is the boat moving after 8
seconds?
What do I
know?
What do I
need?
vo = 6 m/s
v=?
a = +2.0 m/s2
v  v o at
m
v  22
s
t = 8 sec




v  6 ms   2.0 sm2 8sec
Equation #2
1 2
x  v ot  at
2
b. Find the displacement of the boat after the 8 second.
x  v ot 

1 2
at
2
x  6.0 ms 8sec 



1
2
2.0 sm2 8sec
2
x  112 m only 2 significant figures therefore
x  110 m
Graphical Representation
Velocity vs time
Area of triangle
25
22 m/s
20
Velocity (m/s)
1
bh
2
1
At  816
2
At  64 m
At 
15
Velocity vs time
10
Area of rectangle
5

0
0
6 m/s
2
4
6
8
Time (sec)
Total Displacement = 64 m + 48 m
= 112 m  110 m
10
Ar  bh
Ar  86
Ar  48 m
Equation/Example #3
v  vo  2ax
2
2
Example: You are driving through town at 12 m/s when suddenly a ball rolls
out in front of your car. You apply the brakes and begin decelerating at
3.5 m/s2.
How far do you travel before coming to a complete stop?
What do I know?

What do I want?
vo = 12 m/s
x=?
v 2  v o  2ax
2
x
a = - 3.5 m/s2
v = 0 m/s

v2  vo
2a 

0  12 
23.5
2
x
2
x  20.57 m
2
Confusion??
How do I know what equation to use and what variable do I need to solve for?
Equation
v  vo  at
2
1
x  vo t  2 at
v  vo  2ax
2
2
Missing quantity
x
v
t
VERTICAL Kinematics
All of the 3 previous equations can be applied to the vertical
direction simply by adding a few changes.
v  v o  at
x  v o t  at
1
2
v y  v oy  gt
2
v  v o  2ax
2
2
y  v oy t  gt
1
2
2
v y  v oy  2gy
2
2
Notice: Since g has a value of 9.8 m/s2 and is directed downward, the positive
sign has been replaced with a negative sign.