Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Chapter 4: Motion with a Changing
Velocity
•Motion Along a Line
•Graphical Representation of Motion
•Free Fall
•Projectile Motion
•Apparent Weight
•Air Resistance and Terminal Velocity
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§4.1 Motion Along a Line
For constant acceleration the kinematic equations are:
1
x  x f  xi  vix t  a x t 2
2
v x  v fx  vix  a x t
v fx  vix  2a x x
2
2
x  vav, x t
Also:
vav, x 
vix  v fx
2
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: In a previous example, a box sliding across a rough
surface was found to have an acceleration of -2.94 m/s2. If
the initial speed of the box is 10.0 m/s, how long does it take
for the box to come to rest?
Know: a= -2.94 m/s2, vix=10.0 m/s, vfx= 0.0 m/s
Want: t.
v x  vix  a x t  0
vix
 10.0 m/s
t  

 3.40 sec
2
ax
-2.94 m/s
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 4.8): A train of mass 55,200 kg is
traveling along a straight, level track at 26.8 m/s. Suddenly
the engineer sees a truck stalled on the tracks 184 m ahead.
If the maximum possible braking force has magnitude
84.0 kN, can the train be stopped in time?
Know: vfx = 0 m/s, vix=26.8 m/s, x=184 m
Determine ax and compare to the train’s maximum ax.
v x  vix  2a x x  0
2
2
 vix
 1.95 m/s 2
2x
2
ax 
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
The train’s maximum acceleration is:
a x ,max 
Fnet
m

Fbraking
m
 1.52 m/s 2
The maximum acceleration is not sufficient to
stop the train before it hits the stalled truck.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§4.2 Visualizing Motion with
Constant Acceleration
Motion diagrams for three carts:
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Graphs of x, vx, ax for each of the three carts
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 4.13): A trolley car in New Orleans
starts from rest at the St. Charles Street stop and has a
constant acceleration of 1.20 m/s2 for 12.0 seconds.
(a) Draw a graph of vx versus t.
16
14
12
v (m/sec)
10
8
6
4
2
0
0
2
4
8
6
10
12
14
t (sec)
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
(b) How far has the train traveled at the end of the 12.0
seconds?
The area between the curve and the time axis represents
the distance traveled.
1
x  vt  12 sec   t
2
1
 14.4 m/s 12 s   86.4 m
2
(c) What is the speed of the train at the end of the 12.0 s?
This can be read directly from the graph, vx=14.4 m/s.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§4.3 Free Fall
A stone is dropped from the edge of a cliff; if air resistance
can be ignored, the FBD for the stone is:
y
Apply Newton’s Second Law
x
w
F
y
  w  mg  ma
 a   g  9.8 N/kg
 9.8 m/s 2
The stone is in free fall; only the force of gravity acts on
the stone.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: You throw a ball into the air with speed 15.0 m/s;
how high does the ball rise?
y
viy
Given: viy=+15.0 m/s; ay=-9.8 m/s2
x
ay
To calculate the final height, we
need to know the time of flight.
Time of flight from:
1
y  viy t  a y t 2
2
v fy  viy  a y t
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
The ball rises
until vfy= 0.
The height:
v fy  viy  a y t  0
viy
15.0 m/s
t    
 1.53 sec
2
ay
- 9.8 m/s
1
y  viy t  a y t 2
2


1
2
 15.0 m/s 1.53 s    9.8 m/s 2 1.53 s 
2
 11.5 m
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 4.22): A penny is dropped from the
observation deck of the Empire State Building 369 m above
the ground. With what velocity does it strike the ground?
Ignore air resistance.
y
Given: viy=0 m/s, ay=-9.8 m/s2,
y=-369 m
x
ay
Unknown: vyf
369 m
Use:
v fy  viy  2a y y
2
2
 2 a y y
v yf  2a y y
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
v yf  2a y y  2 9.8 m/s 2  369 m  85.0 m/s
(downward)
How long does it take for the penny to strike the ground?
Given: viy=0 m/s, ay=-9.8 m/s2, y=-369 m
Unknown: t
1
1
2
y  viy t  a y t  a y t 2
2
2
2y
 t 
 8.7 sec
ay
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§4.4 Projectile Motion
What is the motion of a struck baseball? Once it leaves
the bat (if air resistance is negligible) only the force of
gravity acts on the baseball.
The baseball has ax = 0 and ay = -g, it moves with
constant velocity along the x-axis and with nonzero,
constant acceleration along the y-axis.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: An object is projected from the origin. The initial
velocity components are vix = 7.07 m/s, and viy = 7.07 m/s.
Determine the x and y position of the object at 0.2 second
intervals for 1.4 seconds. Also plot the results.
1
1
2
y  viy t  a y t  a y t 2
2
2
x  vix t
Since the object starts from the origin, y and x
will represent the location of the object at time t.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
t (sec)
x (meters)
y (meters)
0
0
0
0.2
1.41
1.22
0.4
2.83
2.04
0.6
4.24
2.48
0.8
5.66
2.52
1.0
7.07
2.17
1.2
8.48
1.43
1.4
9.89
0.29
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
This is a plot of the x position (black points) and y position
(red points) of the object as a function of time.
12
x,y (m)
10
8
6
4
2
0
0
0.5
1
1.5
t (sec)
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Example continued:
This is a plot of the y position versus x position for the
object (its trajectory).
3
2.5
y (m)
2
1.5
1
0.5
0
0
2
4
6
8
10
x (m)
The object’s path is a parabola.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 4.36): An arrow is shot into the air
with  = 60° and vi = 20.0 m/s.
(a) What are vx and vy of the arrow when t=3 sec?
y
The components of the initial
velocity are:
vix  vi cos   10.0 m/s
60°
x
At t = 3 sec:
viy  vi sin   17.3 m/s
v fx  vix  ax t  vix  10.0 m/s
v fy  viy  a y t  viy  gt  12.1 m/s
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
(b) What are the x and y components of the displacement of
the arrow during the 3.0 sec interval?
y
r
x
1
rx  x  vix t  a x t 2  vix t  0  30.0 m
2
1
1
2
ry  y  viy t  a y t  viy t  gt 2  7.80 m
2
2
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: How far does the arrow in the previous example
land from where it is released?
The arrow lands when y=0.
Solving for t:
t 
2viy
g
1
y  viy t  gt 2  0
2
 3.53 sec
1
The distance traveled is: x  vix t  a x t 2
2
 vix t  0  35.3 m
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§4.5 Apparent Weight
Stand on a bathroom scale.
FBD:
y
N
Apply Newton’s 2nd Law:
x
w
F
y
 N  w  ma y
N  mg  ma y
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The normal force is the force the scale exerts on you. By
Newton’s 3rd Law this is also the force (magnitude only) you
exert on the scale. A scale will read the normal force.
N  mg  a y 
is what the scale reads.
When ay = 0, N = mg. The scale reads your true weight.
When ay0, N>mg or N<mg.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: A woman of mass 51 kg is standing in an elevator.
If the elevator pushes up on her feet with 408 newtons of
force, what is the acceleration of the elevator?
FBD for
woman:
y
N
Apply Newton’s 2nd Law:
x
w
F
y
 N  w  ma y
N  mg  ma y (1)
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
Given: N = 408 newtons, m = 51 kg, g = 9.8 m/s2
Unknown: ay
Solving (1) for ay:
N  mg
ay 
 1.8 m/s 2
m
The elevator could be (1) traveling upward with
decreasing speed, or (2) traveling downward
with increasing speed.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§4.6 Air Resistance
A stone is dropped from the edge of a cliff; if air resistance
cannot be ignored, the FBD for the stone is:
y
Apply Newton’s Second Law
Fd
x
w
F
y
 Fd  w  ma
Where Fd is the magnitude of the drag
force on the stone. This force is
directed opposite the object’s velocity
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Assume that
Fd  bv 2
b is a parameter that depends on the
size and shape of the object.
Since Fdv2, can the object be in equilibrium?
F
y
 Fd  w  ma
bv 2  mg  0
mg
yes, when v  vt 
b
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 4.61): A paratrooper with a fully
loaded pack has a mass of 120 kg. The force due to air
resistance has a magnitude of Fd = bv2 where
b = 0.14 N s2/m2.
(a) If he/she falls with a speed of 64 m/s, what is the
force of air resistance?


Fd  bv  0.14 N s /m 64 m/s   570 N
2
2
2
2
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
(b) What is the paratrooper’s acceleration?
y
FBD:
Apply Newton’s Second Law and
solve for a.
Fd
x
F
y
w
 Fd  w  ma
Fd  mg
 5.1 m/s 2
m
(c) What is the paratrooper’s terminal speed?
a
F
y
 Fd  w  ma  0
bvt2  mg  0
mg
vt 
 92 m/s
b
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Summary
•The Kinematic Equations
•Graphical Representations of Motion
•Applications of Newton’s Second Law & Kinematics
(free fall, projectiles, accelerated motion, air drag)
•Terminal Velocity
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