Acceleration and Free Fall
Chapter 2.2 and 2.3
What is acceleration?

Acceleration measures the rate of
change in velocity.

Average acceleration = change in
velocity/ time required for change
Units for acceleration
aavg
v

t
m/s m
 2
s
s
Sign is very important!

Acceleration has both direction and
magnitude

A negative value for acceleration does
not always mean an object is
decelerating!!
2-4 Acceleration
Increasing speed and deceleration (decreasing
speed) should not be confused with the
directions of velocity and acceleration:
Speeding up, moving to the right
Slowing down, moving to the left
Slowing down, moving to the right
Speeding up, moving to the left
Fill in the Chart
Initial Velocity Acceleration
Motion
+
+
Speeding up, moving
right/up
-
-
Speeding up, moving
left/down
+
-
Slowing Down
moving right/up
-
+
Slowing Down,
moving left/down
- or +
0
Constant Velocity
0
- or +
Speeding up from
rest
0
0
Remaining at rest
Graph of Velocity vs Time
Question: What does the slope of this graph give you?
Answer: ACCELERATION
Rise = Δv
Run Δt
Vf – VAVG = Δv
tf – ti = Δt

The Kinematic Equations

You are going to loooooove these! 
Motion with constant acceleration

Kinematic Equations

The relationships between displacement,
velocity and constant acceleration are
expressed by equations that apply to any
object moving with constant
acceleration.
Displacement with constant acceleration
1
x  (vi  v f )t
2
Δx = displacement
 Vi = initial velocity
 Vf = final velocity
 Δt = time interval

Example: #1 p.53 in book

A car accelerates uniformly from rest to a
speed of 23.7 km/h in 6.5 s. Find the
distance the car travels during this time.
Δx = displacement= distance= ?
 Vi = initial velocity = rest = 0 km/h
 Vf = final velocity = 23.7 km/h
 Δt = time interval = 6.5 s

Look at final velocity…convert to m/s!!!
Problem Solving

Final velocity
km 1000m
1h
m
23.7
x
x
 6.58
conversion
h
1km 3600s
s


Plug in values and
solve for Δx
1 m
m
x  (0  6.58 )(6.5s )  21m
2 s
s
Velocity with constant uniform acceleration
v f  vi  at
Vf = final velocity
Vi = initial velocity
a = acceleration
Δt = time interval
Example: #2 p.55

An automobile with an initial speed of
4.30 m/s accelerates uniformly at the
rate of 3.0 m/s2. Find the final speed
after 5.0 seconds.
Vf = final velocity=?
Vi = initial velocity = 4.3 m/s
a = acceleration= 3.0 m/s^2
Δt = time interval= 5.0 s
Solve

Plug in values and
solve for Vf
m
m
v f  4.3  (3.0 2 )(5.0s )
s
s
Vf= 19 m/s
Displacement with constant uniform acceleration
1
2
x  vi t  a (t )
2
Δx = displacement
Vi = initial velocity
a = acceleration
Δt = time interval
Example: #2 p.55

An automobile with an initial speed of 4.30 m/s
accelerates uniformly at the rate of 3.0 m/s2.
Find the displacement after 5.0 seconds.
Δx = displacement=??
Vi = initial velocity= 4.30 m/s
a = acceleration= 3.0 m/s^2
Δt = time interval= 5.0 s
Solve!

Plug in values and
solve for
displacement
m
1
m
x  (4.3 )(5.0s )  (3.0 2 )(5.0 s ) 2  59m
s
2
s
Final Velocity after any displacement
v  v  2ax
2
f
2
i
Vf = final velocity
Vi = initial velocity
a = acceleration
Δx = displacement
Example: p.58 #3
A car accelerates uniformly in a straight
line from rest at the rate of 2.3 m/s^2.
What is the speed of the car after it has
traveled 55 m?
 Vf = final velocity=??
 Vi = initial velocity= rest= 0 m/s
 a = acceleration= 2.3 m/s^2
 Δx = displacement= 55 m

Solve
2
m
m
m
2
2
v f  (0 )  2(2.3 2 )(55m)  253 2
s
s
s
2
m
m
v  253 2  v f  16
s
s
2
f
Rearranging

Your problems won’t always be so
straightforward…make sure to rearrange
your equations to solve for the unknown
before plugging in your numbers (with
units!)
Section 2-3 Falling Objects

Free Fall: Neglecting air resistance, all
objects fall with the same constant
acceleration
Acceleration due to gravity
m
g  9.81 2
s
Free Fall Acceleration
•
However, acceleration is a vector.
•
Gravity acts toward the earth (down)
•
Therefore, the acceleration of
objects in free fall near the surface of
the earth is
m
a   g  9.81 2
s
What we see because of air
resistance…
Object falling from rest
Path of a projectile
At top of path
v= 0 m/s
a = -9.81 m/s2
Free Fall Acceleration

At the highest point of an arc, an object has
velocity = 0 m/s, acceleration is still -9.81
m/s2

An object thrown into the air is a freely
m
falling body with vi  0
s
Free Fall Problem p.64 #2

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 paserby on the
sidewalk below have to move out of the way
before the flowerpot hits the ground?
Part. A.


What are we looking for: Vf
What do we know?



Displacement: -25 m
Acceleration: -9.81 m/s2
Vi=0 m/s
What equation should we use??
v
2
f
v
2
i
 2ax
Solve the problem 
v f   v  2ax
2
i
2
m
 m
v f    0   2(9.81 2 )( 25m)
s
 s
m
v f  22.1
s
Part b.

How much time before
the flowerpot hits the
ground?







What do we know?
Displacement= -25.0 m
Acceleration = -9.81
m/s2
V initial= 0
V final = -22.1 m/s
What are we looking
for: Time!
Which equation should
we use??
v f  vi  at
Solve the Problem 
v f  vi  at
v f  vi
a
 t
m
 22.1  0
s
t 
m
 9.81 2
s
t  2.25s