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
Position, Velocity, and Speed
of a Particle
∆x
vx ,avg ≡
∆t
Average Velocity
52 m − 30 m
= 2.2 m/s
Example:
10 s − 0
Average Speed
Average speed of particle (scalar quantity): total distance d
traveled divided by elapsed time ∆t
d
vavg ≡
∆t
75 m
vavg =
= +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:
xF − xA
vx ,avg =
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:
127 m
=
vavg =
50.0 s
2.54 m/s
Instantaneous Velocity and Speed
∆x dx
vx ≡ lim
=
∆t → 0 ∆t
dt
Example 2.3:
Average and Instantaneous Velocity
x = −4t + 2t2
(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
x = −4t + 2t2
(B) Calculate the average velocity during these two time
intervals.
∆xA → B −2 m
vx ,avg ( A → B) =
=
= −2 m/s
1s
∆t
∆xB→ D 8 m
vx ,avg ( B→
=
= =
D)
2s
∆t
+4 m/s
Example 2.3:
Average and Instantaneous Velocity
x = −4t + 2t2
(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
x= xi + vi t
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
Average Acceleration
∆vx vxf − vxi
ax ,avg ≡
=
∆t
t f − ti
Instantaneous Acceleration
∆vx dvx
ax ≡ lim
=
∆t → 0 ∆t
dt
Acceleration vs. Time Graph
Force and Acceleration
The force on an object is proportional to the
acceleration of the object.
Fx ∝ ax
Acceleration
dvx d dx d 2 x
=
ax =
=
2
dt
dt dt dt
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
2
vx = 40 − 5t
(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
vxf − vxi vxB − vxA 20 m/s − 40 m/s
=
ax ,avg = =
2.0 s − 0 s
t f − ti
tB − t A
= −10 m/s 2
Example 2.6:
Average and Instantaneous Acceleration
vx = 40 − 5t2
(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
=lim ( −10t − 5∆t ) =−10t
ax =lim
∆t → 0 ∆t
∆t → 0
ax =
−20 m/s 2
( −10 )( 2.0 ) 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 )
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 (a) 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 )
Carrier Landing
A jet lands on an aircraft carrier at a speed of 140 mi/h ( ≈ 63 m/s).
What is its acceleration (assumed constant) if it stops
in 2.0 s due to an arresting cable that snags the jet and
brings it to a stop?
Example 2.7:
Carrier Landing
A jet lands on an aircraft carrier at a speed of 140 mi/h
( ≈ 63 m/s).
(A) What is its acceleration (assumed constant) if it
stops in 2.0 s due to an arresting cable that snags the jet
and brings it to a stop?
vxf = vxi 0 − 63 m/s
2
=
−32 m/s
ax =≈
t
2.0 s
Example 2.7:
Carrier Landing
(B) If the jet touches down at position xi = 0, what is
its final position?
1
x f =+
xi
vxi + vxf ) t
(
2
1
0 + ( 63 m/s + 0 )( 2.0 s ) =
63 m
=
2
Example 2.8:
Watch Out for the Speed Limit!
You are driving at a constant speed of 45.0 m/s when
you pass a trooper on a motorcycle hidden behind a
billboard. One second after your car passes the
billboard, the trooper sets out from the billboard to
catch you,
accelerating at a
constant rate of
3.00 m/s2. How
long does it take
the trooper to
overtake your
car?
xcar= xB + vx car t
1 2
x f =xi + vxi t + ax t
2
1 2 1 2
0 + ( 0 ) t + ax t =ax t
xtrooper =
2
2
xtrooper = xcar
1 2
ax t= xB + vx car t
2
Example 2.8:
Watch Out for the Speed Limit!
1 2
0
ax t − vx car t − xB =
2
2
vx car ± vx2 car + 2ax xB vx car
vx car 2 xB
t=
=
±
+
2
ax
ax
ax
ax
45.0 m/s
t=
+
2
3.00 m/s
( 45.0 m/s )
2
( 3.00 m/s )
2 2
+
2 ( 45.0 m )
3.00 m/s
2
=31.0 s
Freely Falling Objects
Galileo Galilei
Leaning Tower of Pisa
Quick Quiz 2.8 Part I
What happens to the acceleration of a ball after it is
thrown upward into the air? (Neglect air resistance.)
(a) increases
(b) decreases
(c) increases and then decreases
(d) decreases and then increases
(e) remains the same
Quick Quiz 2.8 Part II
What happens to the speed of a ball after it is thrown
upward into the air? (Neglect air resistance.)
(a) increases
(b) decreases
(c) increases and then decreases
(d) decreases and then increases
(e) remains the same
Example 2.10:
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:
(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
0 − 20.0 m/s
=
t tB=
=
2
−9.80 m/s
ay
=
2.04 s
v yB − v yA
−g
Example 2.10:
(B) Find the maximum height of the stone (above its
initial position).
1 2
ymax = yB = y A + vxAt + a y t
2
yB = 0 + ( 20.0 m/s )( 2.04 s )
1
2
2
20.4 m
+ ( −9.8 m/s ) ( 2.04 s ) =
2
Example 2.10:
(C) Determine the velocity of the stone when it returns
to the height from which it was thrown.
v yC =
v yA + 2a y ( yC − yA )
2
v yC
2
2
( 20.0 m/s ) + 2 ( −9.80 m/s ) ( 0 − 0 )
2
= 400 m 2 /s 2
v yC = −20.0 m/s
2
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=
v yA + a y t
yD
= 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
Kinematic Equations
Derived from Calculus
∆x=
vxn ,avg ∆tn =
∆x
n
∑v
n
tf
xn ,avg
∆=
tn
∆x
∆x =∫ vx ( t ) dt
ti
lim ∑ vxn ,avg ∆tn
∆tn → 0
n
Kinematic Equations
Derived from Calculus
dvx
ax =
⇒ dvx = ax dt
dt
t
vxf − vxi =
∫ ax dt
0
t
vxf − vxi = ax ∫ dt= ax ( t − 0 )= ax t
0
Kinematic Equations
Derived from Calculus
dx
vx =
⇒ dx = vx dt
dt
t
x f − xi =
∫ vx dt
0
v=
vxf= vxi + ax t
x
x f − x=
i
t
∫ ( v + a t ) dt
0
xi
x
t2
= ∫ vxi dt + ax ∫ t dt= vxi ( t − 0 ) + ax − 0
0
0
2
1 2
x f − xi = vxi t + ax t
2
t
t
0
