You are going 25 m/s North on
I-35. You see a cop parked on
the side of the road. What is his
velocity related to you.
A. 25 m/s South
B. 25 m/s North
C. 0 m/s
D. 50 m/s North
Acceleration
• Is a change in velocity over
time.
• Is usually described as
“speeding up” or “slowing
down”
• Is a vector!
• Equation for acceleration:
v
a
t
Signs
• Since displacement, velocity, and acceleration
are all vectors, we need to specify a direction
as well as a magnitude when we describe them.
• To do this numerically, we will use the sign of the
number (+ or -) to tell us the direction.
• Common conventions:
– Anything going to the right or up is designated as positive (+)
– Anything going to the left or down is designated as negative (-)
– Like a graph!
• Every situation and physics problem is different, so the
sign conventions might change, but will always need to
be stated for clarity.
• This means that the sign of a number only tells us the
object’s direction!
Motion Graphs
• Let’s go over the basics.
Position vs. time graphs
(x vs. t)
• Shows the change in position for an object over a period
of time.
• The slope of the line at any point shows the velocity of
the object.
• What is the velocity of this
object from t=1 to t=3 seconds?
– 20 m/s
Position vs. time graphs
(x vs. t)
• A parabolic curve means that the slope (velocity) is not
constant, so the object is accelerating.
Velocity vs. time graphs
(v vs. t)
• Shows the change in velocity for an object over a period
of time.
• The slope of the line at any point shows the acceleration
of the object.
• What is the acceleration
of this object from t=4 to
t=8 seconds?
5 m/s2
Velocity vs. time graphs
(v vs. t)
• The total area under the line tells you the displacement
(change in position) for the object in motion.
• What is the displacement of the object from t=0s to t=8s?
160 m
Acceleration vs. time graphs
(a vs. t)
• These graphs are boring, and will only have a straight
line above the axis, below the axis, or at 0.
• But what does the straight line mean?
– Acceleration is always constant.
• What is the acceleration in
this graph?
– -2m/s2
• What does that mean?
– Object is accelerating to the left.
Instantaneous vs. Average
Velocity
• For objects that are accelerating,
we will talk about basically two
types of velocity that the object
will have:
• Instantaneous Velocity
– The velocity of an object
(ex: a motorcycle) at any one
instant. (The speedometer reading)
• Average Velocity
– This would be if you took the initial and final velocity of the object
and averaged them together. You could speed up and slow down
but overall there would be an average velocity
– It is also the average velocity when you take the total
displacement of the object over a certain amount of time.
– 2 Formulas:
vavg 
v f  vi
2
vavg
Δx d


t
t
Instantaneous vs. Average
Velocity
Instantaneous
x2
x
x1
t
t1
t2
Displacement, x
Average
slope
x
t
Time
Equations for 1D Motion
• Before we get started
on problems, here is
your “toolbox” of
equations to use for
this unit
• These are the ones
we will mainly use
vavg
Δx d


t
t
vavg 
(v f  vi )
v (v f  vi )
a

t
t
v f  vi  at
v f  vi  2ad
2
2
d  vi t  1 at 2
2
2
Sample 1-D Motion Problem
A car starts from rest and accelerates to the right at 2.5 m/s2.
How long does it take the car to reach a velocity of 15 m/s?
** Remember to GUESS!**
Givens:
We know that vi is zero
vi = 0 m/s
because the car starts “from
a = +2.5 m/s2 rest.”
vf = 15 m/s
Unknown:
t=?
Equation:
vf = vi + at
t = (vf – vi)/a
Substitute:
t = (15 – 0)/(2.5)
Solve:
t = 6.0 s
Remember to box your answer
and include units!
Look for an equation that has
the variables vi, a, t, and vf
Rearrange the variables to
solve for t
Practice Problem
An airplane is traveling east with a constant velocity of 25 m/s when it
begins to accelerate uniformly at 3.0 m/s2. How far does the airplane
travel in 45 seconds?
Answer:
4,200 m
Practice Problem
A train moving to the right at 30 m/s begins to accelerate to the left at
7.5 m/s2 how long does it take until its speed reaches 0 m/s?
Answer:
4 seconds
Freely Falling Bodies
• Up to now, we have been
looking at 1D motion that has
been horizontal.
• But what about vertical motion?
You know, things rising and
falling?
• This might sound more
complicated, but it’s actually
easier!
Acceleration Due to
Gravity
• Every object on the earth
experiences a common force:
the force due to gravity.
• This force is always directed
toward the center of the earth
(downward).
• The acceleration due to gravity
is relatively constant near the
Earth’s surface.
g
W
Earth
Gravitational Acceleration
• In a vacuum, all objects fall
with same acceleration.
• Equations for constant
acceleration apply as
usual.
• Near the Earth’s surface:
a = g = 10.0 m/s2
Gravity is always directed downward
Freely Falling Bodies
• So what does this
change?
– When working free-fall
problems, we will still
use the exact same
equations!
– Acceleration (a) will
always be 10.0 m/s2.
v f  vi  at
v f  vi  2ad
2
2
d  vi t  1 at 2
2
Sign convention
•
Since we are dealing a lot with objects
going “up” and going “down,” it can be
confusing when signs should be positive
and when they should be negative
(especially for gravity!).
• Here is a good rule to follow:
– Whatever direction is the initial
direction of motion is the positive
direction.
g
+
• This means that if an object is
falling down, then g would be
+10.0 because down would be
positive.
• However, if an object was thrown
up into the air, g would be -10.0
because up would be positive.
– We will apply this rule for
horizontal motion as well.
g
v
+
+
Sample freefall problem
Timmy drops a rock from the top of a cliff and it falls to the ground below.
If the rock takes 8 seconds to hit the ground, what is the height of the
cliff?
We know v is 0
i
Givens:
vi = 0 m/s
a = g = 10.0
t=8s
Draw a diagram!
because he drops
the rock from rest.
a is 10.0m/s2
because the rock is
m/s2 falling due to gravity.
Substitute:
d = (0)(8) + ½ (10.0)(8)2
Unknown:
d=?
Solve:
Equation:
d = vit + ½ at2
d = 320. m
a
Since the length the
rock falls will be the
same as the height
of the cliff, we are
looking for d.
v
d
+
Practice Problem
A football player kicks a football from the ground and it flies
straight up into the air with a velocity of 12 m/s. After
how much time does the ball come to a stop before
falling back to earth?
Answer:
t = 1.2 seconds
What is the height of the ball at the apex?
Answer:
d = 7.2 meters
Practice Problem
A ball is thrown upward from the edge of a building that is
100 meters tall. The time between the throw and the
object hitting the ground is 6.2 seconds. What is the
initial velocity of the ball?
Answer:
vi = 17.1 m/s
What is the maximum height reached by the ball?
Answer:
d = 14.6 meters