Chapter 2




To simplify the concept of motion,
we will first consider motion that
takes place in one direction.
One example is the motion of a
commuter train on a straight track.
To measure motion, you must first
choose a frame of reference. A
frame of reference is a system for
specifying the precise location of
objects in space and time.
In the train example, any station along
the route.



Displacement is a change in position.
Displacement is not always equal to the
distance traveled.
The SI unit of displacement is the meter, m.
∆ x = xf -xi
Displacement – final position – initial position
Displacement is not always equal to the distance
traveled.
Example: If a gecko starts at an initial position of 20
cm and moves to the 80 cm mark, then retreats back
to the 50 cm mark as its final position, How far has
the gecko traveled? What is its displacement?
The gecko traveled 90 cm, but its displacement is 30 cm.

In this course, right (east) is positive as well as
upward (north) and left (west) is negative as well as
downward (south).
Average velocity is the total displacement
divided by the time interval during which the
displacement occurred.
In SI, the unit of velocity is meters per second
abbreviated as m/s.

Consider a trip to a friend’s house 370 km to
the west (negative direction) along a straight
highway. If you left at 10 AM and arrived at 3
PM, what is your average velocity?
This is your average. You did not travel at
74 km/h at every moment.
During a race on a level ground, Andra
runs with an average velocity of 6.02
m/s to the east. What is Andra’s
dispacement after 137 s?
x=vt
= 6.02m/s • 137s = 825 m




Velocity is not the same as speed.
Velocity describes motion with both direction
and a numerical value (magnitude).
Speed has no direction, only magnitude.
Average speed is equal to the total distance
traveled divided by the time interval.


Consider an object whose position-time graph is
not a straight line, but a curve.
We obtain different average velocities depending
on the time interval. The instantaneous velocity is
the velocity of the object at a specific point in the
object’s path
The instaneous velocity can
be determined by measuring
the slope of the line that is
tangent to that point on the
diatance-vs-time graph.
Practice Problems 2A
 Problems 1-6, Page 44




Acceleration – Rate at which velocity changes
over time
An object accelerates if its speed, direction or
both change.
Acceleration has direction and magnitude.
Acceleration is a vector quantity.
Acceleration has the dimensions of
length divided by time squared.
 SI units are m/s2
 Remember we have (m/s)/s = m/s2
Acceleration
A bus slows down with an average acceleration
of -1.8 m/s2. How long does it take the bus to
slow down from 9.0 m/s to a complete stop?


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 for this acceleration to occur?
A treadmill runs with a velocity of -12m/s and
speeds up at regular intervals during a half hour
workout. After 25 minutes, the treadmill has a
velocity of -6.5 m/s. What is the acceleration of
the treadmill during this period?


Consider a train moving to the right, so that the
displacement and velocity are positive.
The slope of the velocity-time graph is the average
acceleration.
• 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.


When velocity changes by the same amount
during each time interval, acceleration is
constant.
The relationships between displacement, time,
velocity, and constant acceleration are
expressed by the equations shown on the next
slide.
These equations apply to any object moving
with constant acceleration. These equations use
the following symbols:

A racing car reaches a speed of 42 m/s. It
begins a uniform negative acceleration , using
its parachute and braking system, and comes to
a complete rest 5.5 s later. Find the distance
that the car travels during braking.



A car accelerates uniformly from rest to a
speed of 6.6 m/s in 6.5 s. Find the distance
traveled during this time.
A driver in a car at a speed of 21.9 m/s sees
a cat 101 m away on the road. How long
does it take the car to accelerate uniformly to
a stop in exactly 99m?
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?
A plane starting at rest at one end of
the 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 takeoff?
A car starts from rest and travels for
5.0 s with a constant acceleration of 1.5m/s2 . What is the final velocity
of the car? How far does the car
travel in this time interval?
A person pushing a
stroller starts from
rest , uniformly
accelerating at a rate
of 0.500 m/s2. What is
the velocity of the
stroller after it has
traveled 4.75 m?
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 is
2.7m/s2 , how far did it travel during the
acceleration?
A car traveling initially at 7.0 m/s accelerates
uniformly at the rate of 0.80 m/s2 for a
distance of 245 m.
What is the velocity at the end of the
acceleration?
What is the velocity after it accelerates for 125 m?
What is the velocity after it accelerates for 67 m?
Practice Problems 2B-2E
 Pages 49 (1-5), 53 (1-4), 55 (1-4)
and 58 (1-6)
August 2, 1971 – On the moon
With NO air resistance objects will fall at exactly
the same rates
Fall with the same constant acceleration.
Free fall is the motion of a body when only
gravity is acting on it.
The acceleration on an object in free fall is called
the acceleration due to gravity, or free-fall
acceleration.
Free-fall acceleration is denoted with the symbols
ag or g
Free-fall acceleration is the same for all
objects regardless of mass.
We will use the value g = 9.81 m/s2
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
Moving upward: velocity is decreasing,
acceleration is -9.81 m/s2
Moving upward: velocity is decreasing,
acceleration is -9.81 m/s2
Jason hits a volleyball so that
it moves with an initial
velocity of 6.0m/s straight
upward. If the volleyball
starts from 2.0m above the
floor, how long will it be in
the air before it strikes the
floor?
Both ∆t and vf are known. Therefore, first
solve for vf using the equation that does not
require time. Then, the equation for vf that
does not involve time can be used to solve for
∆t .
A flowerpot falls from a windowsill 25.0m
above the sidewalk.
How fast is the flowerpot moving when
it hits the ground?
How much time would you have to move
out of the way before the flowerpot hits
the ground?
A robot drops a camera off a 239m cliff on
mars where free-fall acceleration is -3.7
m/s2
Find the velocity with which the camera
hits the ground.
Find the time required to hit the ground
Practice Problems 2F
 Page 64 Problems 1-4
Mixed Review pp 70-71 (33-41)