Graph this (assume motion along a line)
You walk to the bus stop in the morning
(400 meters away) in 4 minutes. After 1
minute, you realize that you forgot your
clicker. You return to your room quickly, in
2 minutes, but it takes you 1 minute to find
your clicker. You return to the bus stop in 2
minutes just in time.
x [m]
400
0
0
2
4
6
8
10
Next: what is the velocity in each interval
(that’s the slope in each section)?
t [min]
x
v
t
Now draw the velocity versus
time graph
x
Recall that velocity is v 
t
This can be interpreted as the slope
of the x versus t graph
v [meter/min]

200
100
0
-100
-200
0
2
4
6
8
10
t [min]
Checking Understanding
Here is a motion diagram of a car moving along a straight
stretch of road.
If the positive-x direction is to the right, which of the
following velocity-versus-time graphs matches this motion
v diagram?
v
v
v
t
A.
t
B.
t
C.
t
D.
Position and Velocity vs. Time
A particle moves with the velocityversus-time graph shown.
Which graph best illustrates the
position of the particle as a function
of time?
A.
B.
C.
D.
4
Position and Velocity vs. Time
A particle moves with the velocityversus-time graph shown.
Which graph best illustrates the
position of the particle as a function
of time?
A.
B.
C.
D.
5
Acceleration
These three motion diagrams show the motion of a particle
along the x-axis. Rank the accelerations corresponding to
these motion diagrams, from most positive to most
negative. There may be ties.
x
1
2
3
6
Acceleration
These three motion diagrams show the motion of a particle
along the x-axis. Which has positive acceleration? Which is
negative?
x
1
2
3
1<3<2
7
Free fall
Constant downward acceleration:
a = –g = –9.8 m/s2
v(t)  v i  g(t  t i )
g
y(t)  y i  v i (t  t i )  (t  t i ) 2
2
2
2
v f  v i  2g y

8
Free Fall Example with stopping
Passengers on The Giant Drop, a free-fall ride at Six Flags
Great America, sit in cars that are raised to the top of a tower.
The cars are then released for 2.0 s of free fall. A) How fast
are the passengers moving at the end of this speeding up
phase of the ride (you can use g=-10m/s2)? B) If the cars in
which they ride then come to rest in a time of 1.0 s, what is
acceleration (magnitude and direction) of this slowing down
phase of the ride? C) Given these numbers, what is the
minimum
0 possible height of the tower?
0
v
tf1=ti2=2.0
a1=-9.8 m/s2
vf1
3.0=tf2
a2
t[s]
The graph helps
to organize the
information given
in the problem.
Free Fall Example with
stopping
A) During the free falling portion of the motion, the acceleration is -g=-9.8 m/s2:
v f  v i  (g) (t f  t i )
v f  (0 m/s) (10 m/s2 )(2.0 s) = 20 m/s

B) During the stopping portion of the motion, we
can use the same equation to find the unknown
acceleration
v f  v i  a2 (t f  t i )
0 m/s  (20 m/s) a2 (1.0 s)
a2  20 m/s2
Note that the acceleration during the stopping portion is
positive (same sign as the slope on the v versus t graph for
 part of the motion)
this
Free Fall Example with
stopping
y (m)
C) To find the net displacement during the two motions,
we need to separately find the displacements during each
part
0
a
y  v i (t f  t i )  (t f  t i ) 2
2
10 m/s2
y1  (0 m/s)(2.0 s - 0 s) 
(2.0 s - 0 s) 2  20 m
2
20 m/s2
y 2  (20 m/s)(3.0 s - 2.0 s) 
(3.0 s - 2.0 s) 2  10 m
2
y net  y1  y 2  (20 m) +(10 m) = 30 m
-20

-30
Since the ride falls 30 meter, the tower had
better be at least that high (about 100 feet)
t (s)
0
2 3