Storyline
Chapter 2:
Motion in One Dimension
Physics for Scientists and Engineers, 10e
Raymond A. Serway
John W. Jewett, Jr.
Position
position x: location of
particle with respect to
chosen reference point
Displacement
Displacement x of particle:
change in position in a given
time interval
x  x f − xi
Distance and Displacement
Vector quantity requires specification of both direction and magnitude
Scalar quantity has numerical value and no direction
Position, Velocity, and Speed
of a Particle
Quick Quiz 2.1
Which of the following choices best
describes what can be determined
exactly from the table and figure for
the entire 50-s interval?
(a) The distance the car moved.
(b) The displacement of the car.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Quick Quiz 2.1
Which of the following choices best
describes what can be determined
exactly from the table and figure for
the entire 50-s interval?
(a) The distance the car moved.
(b) The displacement of the car.
(c) Both (a) and (b).
(d) Neither (a) nor (b).
Position, Velocity, and Speed
of a Particle
vx ,avg
x

t
Average Velocity
52 m − 30 m
Example:
= 2.2 m/s
10 s − 0
Average Speed
Average speed of particle (scalar quantity): total distance d
traveled divided by elapsed time t
vavg
vavg
d

t
75 m
=
= +1.36 m/s
55.0 s
125 m
average speed =
= 2.27 m/s
55.0 s
Example 2.1: Calculating the Average
Velocity and Speed
Find the displacement, average
velocity, and average speed of
the car in the figure between
positions A and F.
x = xF − xA
= −53 m − 30 m
= −83 m
Example 2.1: Calculating the Average
Velocity and Speed
Average velocity:
vx ,avg
xF − xA
=
tF − tA
−53 m − 30 m
=
50 s − 0 s
−83 m
=
= −1.7 m/s
50 s
Example 2.1: Calculating the Average
Velocity and Speed
Average speed:
vavg
127 m
=
= 2.54 m/s
50.0 s
Instantaneous Velocity and Speed
x dx
vx  lim
=
t → 0 t
dt
Quick Quiz 2.3
Members of the highway patrol are more interested in
(a) your average speed or
(b) your instantaneous speed
as you drive.
Quick Quiz 2.3
Members of the highway patrol are more interested in
(a) your average speed or
(b) your instantaneous speed
as you drive.
Example 2.3:
Average and Instantaneous Velocity
A particle moves along the x axis. Its position varies with
time according to the expression x = −4t + 2t2, where x is in
meters and t is in seconds. The position–time graph for this
motion is shown in the figure.
Because the position of the particle
is given by a mathematical
function, the motion of the particle
is known at all times. Notice that
the particle moves in the negative x
direction for the first second of
motion, is momentarily at rest at the
moment t = 1 s, and moves in the
positive x direction at times t > 1 s.
Example 2.3:
Average and Instantaneous Velocity
(A) Determine the displacement of the particle in the
time intervals t = 0 to t = 1 s and t = 1 s to t = 3 s.
xA → B = x f − xi = xB − xA
2
2



= −4 (1) + 2 (1) − −4 ( 0 ) + 2 ( 0 ) 

 

= −2 m
xB→ D = x f − xi = xD − xB
2
2
=  −4 ( 3) + 2 ( 3)  −  −4 (1) + 2 (1) 

 

= +8 m
Example 2.3:
Average and Instantaneous Velocity
(B) Calculate the average velocity during these two time
intervals.
vx ,avg ( A → B)
vx ,avg ( B→ D )
xA → B −2 m
=
=
= −2 m/s
t
1s
xB→ D 8 m
=
=
= +4 m/s
t
2s
Example 2.3:
Average and Instantaneous Velocity
(C) Find the instantaneous velocity of the particle at
t = 2.5 s.
vx =
10 m − ( −4 m )
3.8 s − 1.5 s
= +6 m/s
Analysis Model: Particle Under
Constant Velocity
Analysis model: represents common situation when
solving physics problems
Analysis Model: Particle Under
Constant Velocity
vx ,avg
x
x

→ vx =
t
t
x = x f − xi → vx =
x f − xi
t
x f = xi + vx t
x f = xi + vx t
( for constant vx )
Analysis Model: Particle Under
Constant Velocity
x
vx =
t
x f = xi + vx t
( for constant vx )
Analysis Model: Particle Under
Constant Velocity
x = xi + vi t
Example 2.4:
Modeling a Runner as a Particle
A kinesiologist is studying the biomechanics of the
human body. (Kinesiology is the study of the movement
of the human body. Notice the connection to the word
kinematics.) She determines the velocity of an
experimental subject while he runs along a straight line
at a constant rate. The kinesiologist starts the stopwatch
at the moment the runner passes a given point and stops
it after the runner has passed another point 20 m away.
The time interval indicated on the stopwatch is 4.0 s.
Example 2.4:
Modeling a Runner as a Particle
(A) What is the runner’s velocity?
x x f − xi 20 m − 0
vx =
=
=
= 5.0 m/s
t
t
4.0 s
Example 2.4:
Modeling a Runner as a Particle
(B) If the runner continues his motion after the
stopwatch is stopped, what is his position after 10 s
have passed?
x f = xi + vx t = 0 = 0 + ( 5.0 m/s )(10 s ) = 50 m/s
Analysis Model: Particle Under
Constant Speed
d
v=
t
d
d 2 r 2 (10.0 m )
v=
→ t = =
=
= 12.6 s
t
v
v
5.00 m/s
Analysis Model:
Particle Under Constant Velocity
x
constant velocity: vx =
t
position as a function of time: x f = xi + vx t
Analysis Model:
Particle Under Constant Speed
d
constant speed: v =
t
Analysis Model Approach to
Problem-Solving
• Conceptualize
• Categorize
• Analyze
• Finalize
Average Acceleration
ax ,avg
vx vxf − vxi

=
t
t f − ti
Instantaneous Acceleration
vx dvx
ax  lim
=
t → 0 t
dt
Acceleration vs. Time Graph
Acceleration
dvx d  dx  d 2 x
ax =
=  = 2
dt
dt  dt  dt
Conceptual Example 2.5: Graphical
Relationships Between x, vx, and ax
The position of an object moving along the x axis varies
with time as in the figure. Graph the velocity versus
time and the acceleration versus time for the object.
Conceptual Example 2.5: Graphical
Relationships Between x, vx, and ax
Conceptual Example 2.5: Graphical
Relationships Between x, vx, and ax
Example 2.6:
Average and Instantaneous Acceleration
The velocity of a particle moving
along the x axis varies according
to the expression vx = 40 − 5t2,
where vx is in meters per second
and t is in seconds.
Example 2.6:
Average and Instantaneous Acceleration
(A) Find the average acceleration in
the time interval t = 0 to t = 2.0 s.
vxA = 40 − 5tA 2 = 40 − 5 ( 0 ) = +40 m/s
2
vxB = 40 − 5tB 2 = 40 − 5 ( 2.0 ) = +20 m/s
2
ax ,avg =
vxf − vxi
t f − ti
=
= −10 m/s 2
vxB − vxA 20 m/s − 40 m/s
=
tB − t A
2.0 s − 0 s
Example 2.6:
Average and Instantaneous Acceleration
(B) Determine the acceleration
at t = 2.0 s.
vxf = 40 − 5 ( t + t ) = 40 − 5t − 10t t − 5 ( t )
2
2
vx = vxf − vxi = −10t t − 5 ( t )
2
vx
ax = lim
= lim ( −10t − 5t ) = −10t
t → 0 t
t → 0
ax = ( −10 )( 2.0 ) m/s 2 = −20 m/s 2
2
Motion Diagrams
Quick Quiz 2.6
Which one of the following statements is true?
(a) If a car is traveling eastward, its acceleration must
be eastward.
(b) If a car is slowing down, its acceleration must be
negative.
(c) A particle with constant acceleration can never stop
and stay stopped.
Quick Quiz 2.6
Which one of the following statements is true?
(a) If a car is traveling eastward, its acceleration must
be eastward.
(b) If a car is slowing down, its acceleration must be
negative.
(c) A particle with constant acceleration can never
stop and stay stopped.
Analysis Model: Particle
Under Constant Acceleration
ax =
vxf − vxi
t −0
vxf = vxi + ax t ( for constant ax )
vx ,avg =
vxi + vxf
2
( for constant ax )
Analysis Model: Particle
Under Constant Acceleration
vxi + vxf
x
vx ,avg = , vx ,avg =
t
2
x = x f − xi , t = t f − ti = t − 0 = t
1
x f − xi = vx , avg t = ( vxi + vxf ) t
2
1
x f = xi + ( vxi + vxf ) t
2
( for constant ax )
Analysis Model: Particle
Under Constant Acceleration
1
vxf = vxi + ax t substitute into x f = xi + ( vxi + vxf ) t
2
1
x f = xi + vxi + ( vxi + ax t )  t
2
1 2
x f = xi + vxi t + ax t
2
( for constant ax )
Analysis Model: Particle
Under Constant Acceleration
vxf = vxi + ax t → t =
vxf − vxi
ax
1
substitute into x f = xi + ( vxi + vxf ) t
2
 vxf − vxi
1
x f = xi + ( vxi + vxf ) 
2
 ax
vxf2 = vxi2 +2ax ( x f − xi )

 = xi +

vxf2 − vxi2
2ax
( for constant ax )
Analysis Model: Particle
Under Constant Acceleration
vxf = vxi + ax t
1 2
x f = xi + vx t + ax t
2
vxf = vxi = vx 
 when ax = 0
x f = xi + vx t 
Kinematic Equations
vxf = vxi + ax t
vx ,avg =
vxi + vxf
2
1
x f = xi + ( vxi + vxf ) t
2
1 2
x f = xi + vxi t + ax t
2
vxf 2 = vxi 2 + 2ax ( x f − xi )
( 2.13)
( 2.14 )
( 2.16 )
( 2.15 )
( 2.17 )
Constant acceleration only!
Quick Quiz 2.7
In the figure, match each vx–t graph on the top with the
ax–t graph on the bottom that best describes the motion.
Quick Quiz 2.7
In the figure, match each vx–t graph on the top with the
ax–t graph on the bottom that best describes the motion.
(a)–(e), (b)–(d), (c)–(f)
Analysis Model:
Particle Under Constant Acceleration
vxf = vxi + ax t
vx ,avg =
vxi + vxf
2
1
x f = xi + ( vxi + vxf ) t
2
1 2
x f = xi + vxi t + ax t
2
vxf 2 = vxi 2 + 2ax ( x f − xi )
Example 2.10:
Not a Bad Throw for a Rookie!
A stone thrown from the top of
a building is given an initial
velocity of 20.0 m/s straight
upward. The stone is launched
50.0 m above the ground, and
the stone just misses the edge
of the roof on its way down as
shown in the figure.
Example 2.10:
Not a Bad Throw for a Rookie!
(A) Using tA = 0 as the time the stone leaves the
thrower’s hand at position A, determine the time at
which the stone reaches its maximum height.
v yf = v yi + a y t  t =
v yf − v yi
ay
=
v yB − v yA
−g
0 − 20.0 m/s
t = tB =
= 2.04 s
2
−9.80 m/s
Example 2.10:
Not a Bad Throw for a Rookie!
(B) Find the maximum height of the stone (above its
initial position).
ymax
1 2
= yB = y A + vxAt + a y t
2
yB = 0 + ( 20.0 m/s )( 2.04 s )
1
2
2
+ ( −9.8 m/s ) ( 2.04 s ) = 20.4 m
2
Example 2.10:
Not a Bad Throw for a Rookie!
(C) Determine the velocity of the stone when it returns
to the height from which it was thrown.
v yC 2 = v yA 2 + 2a y ( yC − yA )
v yC = ( 20.0 m/s ) + 2 ( −9.80 m/s 2 ) ( 0 − 0 )
2
2
= 400 m /s
2
2
v yC = −20.0 m/s
Example 2.10:
Not a Bad Throw for a Rookie!
(D) Find the velocity and position of the stone at
t = 5.00 s.
v yD = v yA + a y t
= 20.0 m/s + ( −9.80 m/s 2 ) ( 5.00 s ) = −29.0 m/s
1 2
yD = yA + v yA t + a y t
2
1
2
2
= 0 + ( 20.0 m/s )( 5.00 s ) + ( −9.80 m/s ) ( 5.00 s )
2
= −22.5 m
Example 2.10:
Not a Bad Throw for a Rookie!
What if the throw were from 30.0 m above the ground
instead of 50.0 m? Which answers in parts (A) to (D)
would change?
None of the answers would change.