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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Chapter 3: Acceleration and
Newton’s Second Law of Motion
•Position & Displacement
•Speed & Velocity
•Acceleration
•Newton’s Second Law
•Relative Velocity
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§3.1 Position & Displacement
The position (r) of an object describes its location relative to
some origin or other reference point.
The displacement is the change in an object’s position. It
depends only on the beginning and ending positions.
r  r f  ri
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 3.4): Margaret walks to the store
using the following path: 0.500 miles west, 0.200 miles north,
0.300 miles east. What is her total displacement? Give the
magnitude and direction.
y
Take north to be in
the +y direction and
east to be along +x.
r3
r
r2
r1
x
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
The displacement is r = rf – ri. The initial position is the
origin; what is rf?
The final position will be rf = r1 + r2 + r3. The components
are rfx = -r1 + r3 = -0.2 miles and rfy = +r2 = +0.2 miles.
y
ry
Using the figure, the magnitude and
direction of the displacement are
r

rx
x
r  rx2  ry2  0.283 miles
tan  
ry
rx
 1 and   45 N of W.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§3.2 Velocity
Velocity is a vector that measures how fast and in what
direction something moves.
Speed is the magnitude of the velocity. It is a scalar.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
r
Average velocity  v av 
t
x 

The
x
component
would
be
:
v



av, x
t 

finish
vav is the constant speed
that results in the same
displacement in a given
time interval.
Path of a
particle
r
Start
Average speed 
distance traveled
time of trip
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
A particle moves along the blue path as shown. At time t1 its
position is r0 and at time t2 its position is rf.
y
v0
r
r0
vf
The instantaneous
velocity points
tangent to the path.
rf
x
r
v av 
t
Points in the direction of r
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
On a graph of position versus time, the average velocity is
represented by the slope of a chord.
x (m)
x2
x1
t1
t2
Average velocity  vav, x
t (sec)
x2  x1

t 2  t1
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
r
Instantane ous velocity  v  lim
t 0 t
This is represented by the slope of a line tangent to the curve
on the graph of an object’s position versus time.
x (m)
t (sec)
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The area under a velocity versus time graph (between the
curve and the time axis) gives the displacement in a given
interval of time.
v(m/s)
t (sec)
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: Consider Margaret’s walk to the store in the
example on slides 3 and 4. If the first leg of her walk takes 10
minutes, the second takes 8 minutes, and the third 7 minutes,
compute her average velocity and average speed during each
leg and for the overall trip.
Use the definitions:
r
Average velocity  v av 
t
distance traveled
Average speed 
time of trip
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
Leg
t
(hours)
vav
(miles/hour)
Average
speed
(miles/hour)
1
0.167
3.00 (west)
3.00
2
0.133
1.50 (north)
1.50
3
0.117
2.56 (east)
2.56
Total
trip
0.417
0.679
2.40
(45 N of W)
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 3.20): Speedometer readings are
obtained and graphed as a car comes to a stop along a
straight-line path. How far does the car move between
t = 0 and t =16 seconds?
Since there is not a reversal of direction, the area
between the curve and the time axis will represent the
distance traveled.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
The rectangular portion has an area of Lw = (20 m/s)(4 s)
= 80 m.
The triangular portion has an area of ½bh = ½(8 s) (20 m/s)
= 80 m.
Thus, the total area is 160 m. This is the distance traveled
by the car.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§3.3 Newton’s Second Law of
Motion
A nonzero acceleration changes an object’s state of motion.
v
Average accelerati on  a av 
t
v
Instantane ous accelerati on  a  lim
t 0 t
These have
interpretations
similar to vav
and v.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
A particle moves along the blue path as shown. At time t1 its
position is r0 and at time t2 its position is rf.
y
v0
v
a av 
t
Points in the
direction of v.
v
vf
r0
rf
x
The instantaneous acceleration
can point in any direction.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 3.31): If a car traveling at 28 m/s is
brought to a full stop 4.0 s after the brakes are applied, find
the average acceleration during braking.
Given: vi = +28 m/s, vf = 0 m/s, and t = 4.0 s.
aav 
v 0  28 m/s

 7.0 m/s 2
t
4.0 s
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 3.39): At the beginning of a 3 hour trip
you are traveling due north at 192 km/hour. At the end, you
are traveling 240 km/hour at 45 west of north.
(a) Draw the initial and final velocity vectors.
y (north)
vf
v0
x (east)
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
(b) Find v.
The components are
vx  v fx  vox  v f sin 45  0  170 km/hr
v y  v fy  voy  v f cos 45  v0  22.3 km/hr
v  vx  v y  171 km/hr
2
tan  
v y
v x
2
 0.1312    tan 1 0.1312   7.5
South of
west
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
(c) What is aav during the trip?
a x ,av
v
a av 
t
a y ,av
v x  170 km/hr


 56.7 km/hr 2
t
3 hr
v y  22.3 km/hr


 7.43 km/hr 2
t
3 hr
The magnitude and direction are:
a av  ax2,av  a y2,av  57.2 km/hr 2
tan  
a y ,av
a x ,av
 0.1310    tan 1 (0.1310)  7.5 South of west
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Newton’s 2nd Law:
The acceleration of a body is directly proportional to the net
force acting on the body and inversely proportional to the
body’s mass.
Mathematically:
Fnet
a
or Fnet  ma
m
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
An object’s mass is a measure of its inertia. The more mass,
the more force is required to obtain a given acceleration.
The net force is just the vector sum of all of the forces acting
on the body, often written as F.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
If a = 0, then F = 0. This body can have:
Speed = 0 which is called static equilibrium, or
speed  0, but constant, which is called dynamic
equilibrium.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§3.4 Applying Newton’s Second Law
F  ma
Force units: 1 N = 1 kg m/s2.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: Find the tension in the cord connecting the two
blocks as shown. A force of 10.0 N is applied to the right on
block 1. Assume a frictionless surface. The masses are m1
= 3.00 kg and m2 = 1.00 kg.
block 2
block 1
F
Assume that the rope stays taut so that both blocks
have the same acceleration.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
FBD for block 2:
FBD for block 1:
y
y
N1
N2
F
T
T
x
x
w2
w1
Apply Newton’s 2nd Law to each
block:
 Fx  F  T  m1a
 Fx  T  m2 a
F
y
 N 2  w2  0
F
y
 N1  w1  0
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
F  T  m1a (1)
T  m2 a (2)
These two equations contain the
unknowns: a and T.
To solve for T, a must be eliminated. Solve for a in (2) and
substitute in (1).
T 
F  T  m1a  m1  
 m2 
T 
 m1 


T
F  m1    T  1 
 m2 
 m2 
F
10 N
T 

 2 .5 N
 m1   3 kg 
1 
 1 

 m2   1 kg 
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: A box slides across a rough surface. If the
coefficient of kinetic friction is 0.3, what is the acceleration of
the box?
y
FBD for
box:
Apply Newton’s 2nd Law:
N
Fk
x
F
F
x
  Fk  ma
y
 N w0
w
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
(1)
 Fk  ma
(2)
N  w  0  N  w  mg
From (1):
 Fk  k N  k mg  ma
Solving for a:


a    k g  0.3 9.8 m/s 2  2.94 m/s 2
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§3.5 Relative Velocity
Example: You are traveling in a car (A) at 60 miles/hour east
on a long straight road. The car (B) next to you is traveling
at 65 miles/hour east. What is the speed of car B relative to
car A?
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
t=0
A
B
+x
t>0
rAG
rBA
A
B
rBG
From the picture: rBG  rAG  rBA
rBA  rBG  rAG
Divide by t: v BA  v BG  v AG
v BA  65 miles/hr east - 60 miles/hr east
 5 miles/hour east
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: You are traveling in a car (A) at 60 miles/hour
east on a long straight road. The car (B) next to you is
traveling at 65 miles/hour west. What is the speed of car B
relative to car A?
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
+x
t=0
t>0
B
From the picture:
Divide by t:
rBG
t>0
A
rAG
B
rBA
A
rBA  rBG  rAG
v BA  v BG  v AG
 65 miles/hr w est - 60 miles/hr east
 125 miles/hr w est
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Summary
•Position
•Displacement Versus Distance
•Velocity Versus Speed
•Acceleration
•Instantaneous Values Versus Average Values
•Newton’s Second Law
•Relative Velocity
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Additional clicker question:
A ball incident on a wall has a speed of 10 m/s toward the
wall. It rebounds with a speed of 10 m/s. What is the
direction of the ball’s acceleration while it is in contact
with the wall?
A. Toward the wall
B. Away from the wall
C. Up the wall
D. Down the wall
E. The ball is not accelerated.
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