13.2 Straight-Line Motion
• Acceleration specified as a function of velocity
Aerodynamic and hydrodynamic forces can cause
an object’s acceleration to depend on its velocity
(Fig 13.13), thus
dv
 a(v)
(13.14)
dt
Because av  is not known as a
function of time, hence we have to
use the separable variables method.
1
13.2 Straight-Line Motion
Using the separable variables method
From Eqn (13.14),
dv
 dt
a (v )
(13.15)
and integrating to obtain,
t
dv
v0 av   t0dt
v
13.16
2
13.2 Straight-Line Motion
In principle, we can solve for Eqn (13.16) for the
velocity as a function of time and then integrate
the relation
ds
v
dt
to determine the position as a function of time.
By using Chain rule, we also can determine the
velocity as a function of the position.
3
13.2 Straight-Line Motion
By writing the acceleration as
dv dv ds dv

 v
dt ds dt ds
And substituting it into Eqn (13.14), we obtain
dv
v  av 
ds
4
13.2 Straight-Line Motion
Separating variables yields
v dv
 ds
av 
and by integration, we can obtain a relation
between the velocity and position. (See example
13.4)
v v dv
s
v0 av   s0ds
5
13.2 Straight-Line Motion
• Acceleration specified as a function of position
Gravitational forces and forces exerted by springs
can cause an object’s acceleration to depend on
its position. If the acceleration is a known function
of position such that
dv
 as 
dt
13.17
6
13.2 Straight-Line Motion
Then we cannot integrate w.r.t time to determine
the velocity because a(s) is not known as a
function of time. Separable variables cannot be
used because there are 3 variables: v, t and s.
However by using Chain rule,
dv dv ds dv

 v
dt ds dt ds
7
13.2 Straight-Line Motion
Rewriting Eqn (13.17) as
dv
v  as 
ds
We can separate the variables,
v dv  as  ds
13.18
and we integrate it to become:
v
s
v0 v dv  s0 as  ds
13.19
8
13.2 Straight-Line Motion
Solving Eqn (13.19) for the velocity as a function
of position becomes,
ds
v   v s 
dt
13.20
then separate the variables and integrate to
determine the position as a function of time:
t
ds
s0 vs   t0dt
s
9
13.2 Straight-Line Motion
If you know a = a(v)
Separate variables:
dv
 av 
dt
dv
 dt
a v 
(or applying the Chain
Rule)
dv dv ds dv

 v  a 
dt ds dt ds
v dv
 ds
av 
(and then separate the
variables)
10
13.2 Straight-Line Motion
If you know a = a(s)
(and then separate the
variables)
Apply the Chain Rule:
dv dv ds dv

 v  as 
dt ds dt ds
v dv  as  ds
*Above Table 13.1 summarizes the procedures
used to determine the velocity when the
acceleration is a known function of velocity or
position.
11
Example 13.4 Acceleration that is a
Function of Velocity
 Question
After deploying its drag parachute, the
airplane shown in Fig 13.14 has an
acceleration a = -0.004v2 m/s2, where v is the
airplane’s velocity in m/s.
a) Determine the time required for the velocity to
decrease from 80 m/s to 10 m/s.
b) What distance does the plane cover during
that time?
12
Example 13.4 Acceleration that is a
Function of Velocity
 Strategy
a) The airplane’s acceleration is known as a
function of its velocity. We can separate the
variables as in Table 13.1 and integrate to
determine the velocity as a function of time.
13
Example 13.4 Acceleration that is a
Function of Velocity
b) We will first use the chain rule to express the
acceleration in terms of a derivative w.r.t the
position s, then separate variables and
integrate to determine the velocity as a
function of s.
14
Example 13.4 Acceleration that is a
Function of Velocity
 Solution
a) The acceleration is
dv
a   0.004v 2
dt
Separating the variables,
dv
v
2
 0.004 dt
15
Example 13.4 Acceleration that is a
Function of Velocity
And integrate, defining t = 0 to be the time at
which v = 80 m/s:
v
dv
t
80 v2  0  0.004 dt
v
 1   0.004t t
0
 v 
80
1 1
   0.004 (t  0)
v 80
16
Example 13.4 Acceleration that is a
Function of Velocity
Solving for t, we obtain
1 1

t  250  
 v 80 
From this eqn, we find that the time required for
the plane to slow to v = 10 m/s is 21.9 s.
Fig 13.15 shows the
velocity of the airplane
as a function of time.
17
Example 13.4 Acceleration that is a
Function of Velocity
b) We write the acceleration as
dv dv ds dv
a 
 v  0.004v 2
dt ds dt ds
Separating the variables yields
dv
 0.004 ds
v
We can now integrate, defining s = 0 to be the
position at which v = 80 m/s:
18
Example 13.4 Acceleration that is a
Function of Velocity
v
80
s
dv
   0.004 ds
0
v
ln v
v
80
 0.004s 
s
0
ln v  ln 80  0.004 ( s  0)
80 

Solving for s, we obtain s  250 ln   m
v
and the distance required to slow to v = 10 m/s is
520 m.
19
Example 13.5 Gravitational (PositionDependent) Acceleration
 Question
In terms of the distance s from the center of the
earth, the magnitude of the acceleration due to
gravity is gRE2/s2, where RE is the radius of the
earth. If a spacecraft is a distance s0 from the
center of the earth. What outward velocity v0 must
it be given to reach a specified distance h from
the earth’s center?
20
Example 13.5 Gravitational (PositionDependent) Acceleration
 Strategy
Acceleration is known as a function of the
position s. we can apply the Chain Rule and
separate variables as shown in Table 13.1,
then integrate to determine the velocity as a
function of s.
21
Example 13.5 Gravitational (PositionDependent) Acceleration
 Solution
The acceleration due to gravity is toward the
center of the earth:
gRE2
a 2
s
Applying Chain Rule results in
dv dv ds dv
gRE2
a

 v 2
dt ds dt ds
s
22
Example 13.5 Gravitational (PositionDependent) Acceleration
Using initial conditions (v = v0 when s = s0) as the
lower limits and the final condition (v = 0 when s =
h) as the upper limits integrating
2
0
h gRE
v0 v dv   s0 s 2 ds
2 0
h
v
2 1 
   gRE  
 s  s0
 2  v0
2
v0
0
2
2 1
 gRE ( 
h
1
)
s0
23
Example 13.5 Gravitational (PositionDependent) Acceleration
Solving for v0, we obtain the initial velocity
necessary for the spacecraft to reach a distance
h:
v0 
2gRE2
 1 1
  
 s0 h 
24
13.3 Curvilinear Motion
•Cartesian Coordinates
let r be the position vector of a point P relative to
the origin O of a cartesian reference frame (Fig
13.17).
The components of r
are the x, y and z
coordinates of P:
r  xi  yj  zk
25
13.3 Curvilinear Motion
Hence the velocity of P relative to the reference
frame is
dr dx dy dz
13.21
v   i  j k
dt dt dt
dt
Expressing in terms of scalar components yields:
v  v xi  v y j  v z k
where
dx
dy
dz
vx  , v y  , vz 
dt
dt
dt
13.22
13.23
26
13.3 Curvilinear Motion
The acceleration of P is
dv dvx dvy
dvz
a

i
j
k
dt
dt
dt
dt
Expressing in terms of scalar components yields:
a  a xi  a y j  a z k
13.24
dvy
dvx
dvz
, ay 
, az 
where ax 
dt
dt
dt
13.25
27
13.3 Curvilinear Motion
Eqns (13.23) & (13.25) describes the motion of a
point relative to a cartesian coordinate system,
which is similar to straight-line motion.
For example a projectile problem. An object is
thrown through the air with negligible drag. It
accelerates due to gravity. In terms of a fixed
cartesian coordinate system with its y-axis
upward, ax = 0, ay = -g and az = 0. At t = 0, the
projectile is located at the origin and has velocity
v0 in the x-y plane at an angle θ0. (Fig 13.18)
28
13.3 Curvilinear Motion
At t = 0, x = 0 and vx = v0cos θ0.
The acceleration in the xdirection is zero i.e.
dvx
ax 
0
dt
Therefore vx is constant and remains equal to its
initial value:
dx
13.26
vx   v0 cos 0
dt
29
13.3 Curvilinear Motion
Integrating Eqn (13.26), we have
x
t
0 dx  0 v0 cos0 dt
x  v0 cos  0  t
13.27 
Thus, we have determined the position and
velocity of the projectile in the x-direction as
functions of time w/o considering the projectile’s
motion in the y- or z-direction.
30
13.3 Curvilinear Motion
At t = 0, y = 0 and vy = v0sin θ0. The acceleration
in the y-direction is
dvy
ay 
 g
dt
By integrating, we obtain:
vy
t
v0 sin 0 dvy  0  g dt
dy
vy 
 v0 sin  0  gt
dt
13.28
31
13.3 Curvilinear Motion
Integrating Eqn (13.28) yields:
y
0
t
dy   (v0 sin  0  gt ) dt
0
1 2
13.29
y  (v0 sin  0 ) t  gt
2
From this analysis, the same vertical velocity and
position are obtained by throwing the projectile
straight up with initial velocity v0sin θ0 (Fig 13.19
a, b). The vertical motion is independent of the
horizontal motion.
32
13.3 Curvilinear Motion
33
13.3 Curvilinear Motion
By solving Eqn (13.27) for t and substituting the
result in Eqn (13.29), we can obtain an equation
describing the parabolic trajectory of the projectile:
y  tan 0  x 
g
2v02
cos 0
2
x2
13.30
34
Example 13.6 Analysis of Motion in
terms of Cartesian Components
 Question
During a test flight in which a helicopter starts
from rest at t = 0 at the origin of the coordinate
system shown in Fig 13.20, accelerometers
mounted onboard the
craft indicate that its
components of acceleration
from t = 0 to t = 10 s are
closely approximated by
35
Example 13.6 Analysis of Motion in
terms of Cartesian Components
a x  0.6 t m/s 2
a y  1.8  0.36 t m/s 2
az  0
Determine the helicopter’s velocity and position
as functions of time.
36
Example 13.6 Analysis of Motion in
terms of Cartesian Components
 Strategy
We can analyze the motion in each coordinate
direction independently, integrating the
acceleration to determine the velocity and then
integrating the velocity to determine the
position.
37
Example 13.6 Analysis of Motion in
terms of Cartesian Components
 Solution
Given at t = 0, v = 0 and we assume that
x = y = z = 0 at t = 0. The acceleration in the xdirection is
dvx
ax 
 0.6 t m/s 2
dt
Integrating w.r.t time yields:
vx
0
t
dvx   0.6 t dt
0
38
Example 13.6 Analysis of Motion in
terms of Cartesian Components
And we obtain the velocity component vx as a
function of time:
dx
vx   0.3t 2 m/s
dt
Integrating again results in
x
0
t
dx   0.3 t 2 dt
0
x  0.1t 3 m
39
Example 13.6 Analysis of Motion in
terms of Cartesian Components
For the y-direction, the acceleration is
ay 
dvy
dt
 1.8  0.36 t m/s 2
Integrating it we have,
vy
0
t
dvy   (1.8  0.36 t ) dt
0
dy
2
vy 
 1.8 t  0.18 t m/s
dt
40
Example 13.6 Analysis of Motion in
terms of Cartesian Components
Integrating the velocity component determine the
position,
y
t
2
dy

(
1
.
8
t

0
.
18
t
) dt


0
0
y  0.9 t 2  0.06 t 3 m
Hence it can be
easily shown that
vz = 0 and z = 0
41
Example 13.7 A Projectile Problem
 Question
The skier in Fig 13.21 leaves the 20° surface at
10 m/s.
a) Determine the distance d to the point where he
lands.
b)What are the magnitudes of
his components of velocity
parallel and perpendicular to
the 45° surface just before he
lands?
42
Example 13.7 A Projectile Problem
 Strategy
a) By neglecting aerodynamic drag and
treating the skier as a projectile, we can
determine his velocity and position as
functions of time.
Using the eqn describing the straight surface
on which he lands, we can relate his
horizontal and vertical coordinates at impact
and thereby obtain an eqn for the time at
which he lands.
Knowing the time, we can determine his
position and velocity.
43
Example 13.7 A Projectile Problem
b) We can determine his velocity parallel and
perpendicular to the 45° surface by using the
result that the component of a vector U in the
direction of a unit vector e is (e.U)e.
44
Example 13.7 A Projectile Problem
 Solution
In Fig (a) we introduce a coordinate system
with its origin where the skier leaves the
surface. His components of velocity at that
instant (t = 0) are:
vx  10 cos 20  9.40 m/s
v y  10 sin 20  3.42 m/s
45
Example 13.7 A Projectile Problem
The x-component of acceleration is zero, so vx is
constant and the skier’s x-coordinate as a function
of time is,
x  9.40 t m
The y-component of acceleration is
ay 
dvy
dt
 9.81 m/s 2
46
Example 13.7 A Projectile Problem
Integrating to determine velocity vy as a function
of time, we obtain
vy
t
3.42 dvy  0  9.81 dt
dy
vy 
 3.42  9.81t m/s
dt
47
Example 13.7 A Projectile Problem
Integrating again to determine the y-coordinate as
a function of time, we have
y
t
0 dy  0  3.42  9.81t  dt
y  3.42 t  4.905 t 2 m
The slope of the surface is -1, so the linear eqn is
y = (-1)x + A, where A is a constant. At x = 0, the
y-coordinate is -3 m, so A = -3.
48
Example 13.7 A Projectile Problem
Hence the eqn describing the 45° surface is
y  x  3 m
Substituting eqns for x & y,
 3.42 t  4.905t  9.40 t  3
2
Solving for t, we get t = 1.60 s.
49
Example 13.7 A Projectile Problem
Therefore the landing coordinates are:
x  9.40(1.60)  15.0 m
y   3.421.60  4.9051.60  18.0 m
2
And the distance d is
d
15.02  18.0  32  21.3 m
50
Example 13.7 A Projectile Problem
b) The components of the skier’s velocity just
before he lands are
vx  9.40 m/s
v y  3.42  9.81(1.60)  19.1 m/s
and the magnitude of his velocity is
v
9.402   19.12  21.3 m/s
51
Example 13.7 A Projectile Problem
Let e be the unit vector that is parallel to the slope
on which he lands (refer to Fig (a)).
e  cos 45 i  sin 45 j
The component of the velocity parallel to the
surface is
e.v e  cos 45 i  sin 45 j 9.40 i  19.1 je
 20.2 e m/s
52
Example 13.7 A Projectile Problem
The magnitude of the skier’s velocity parallel to
the surface is 20.2 m/s. therefore, the magnitude
of his velocity perpendicular to the surface is
v  20.22  6.88 m/s
2
53
13.3 Curvilinear Motion
• Angular Motion
1. Angular Motion of a Line
From Fig 13.22, the angular velocity of L
relative to L0 is defined by
d
rad/s 

dt
13.31
54
13.3 Curvilinear Motion
The angular acceleration of L relative to L0 is
defined by

d d 2

 2 rad/s 2
dt dt

13.32
Table 13.2 relates the analogy between Eqns
(13.31) & (13.32) and the eqns used to a point
along a straight line. Hence, problems involving
angular motion of a line can be analyzed in a
similar method applied to straight-line motion.
55
13.3 Curvilinear Motion
Straight-Line Motion
Angular Motion
ds
v
dt
d

dt
dv d 2 s
a
 2
dt dt
d d 2

 2
dt dt
56
13.3: Curvilinear Motion
2. Rotating Unit vector
We can describe the angular motion of a unit
vector e in a plane just as we described the
angular motion of a line. From Fig 13.23(a), the
direction of L relative to a reference line L0 is
specified by the angle θ and the rate of rotation of
e relative to L0 is specified by the angular velocity
d

dt
57
13.3 Curvilinear Motion
The time derivative of e is defined by
de
et  t   et 
 lim
dt t 0
t
Fig 13.23(b) shows the
vector e at time t and at
time t + Δt. The change in
e during this interval is Δe
= e(t + Δt) – e(t).
58
13.3 Curvilinear Motion
The angle through which e rotates is Δθ = θ(t +
Δt) - θ(t). Since the triangle in Fig 13.23(b) is
isosceles, so
 
 


e  2 e sin 
  2 sin 

 2 
 2 
Introducing a unit vector n that points in the
direction of Δe,
 

e  e n  2 sin 
n
 2 
59
13.3 Curvilinear Motion
The time derivative of e is
 

2 sin 
n

de
e
 2 
 lim
 lim
dt t 0 t t 0
t
 

sin 
 
de
2


 lim
n
dt t 0 
t
2
60
13.3 Curvilinear Motion


sin 
 dθ
2
As t  0,
 1 and


t dt
2
From Fig 13.23(c), the unit vector n is perpendicular
to e(t). Therefore the time derivative of e is
de d

n n
dt dt
13.33
61
13.3 Curvilinear Motion
From Fig 13.23(d), the unit vector n is
perpendicular to e and points in the positive θ
direction.
62
Example 13.8 Analysis of Angular
Motion
 Question
The rotor of a jet engine is rotating at 10,000
rpm when the fuel is shut off. The ensuing
angular acceleration is α = -0.02ω, where ω
is the angular velocity in rad/s.
a) How long does it take the rotor to slow to
1000 rpm?
b) How many revolutions does the rotor turn
while decelerating to 1000 rpm?
63
Example 13.8 Analysis of Angular
Motion
 Strategy
To analyze the angular motion of the rotor, we
define a line L that is fixed to the rotor and
perpendicular to its axis (Fig 13.24). Then we
examine the motion of L relative to the
reference line L0. The angular position, velocity
and acceleration of L describe the angular
motion of the rotor.
64
Example 13.8 Analysis of Angular
Motion
65
Example 13.8 Analysis of Angular
Motion
 Solution
The conversion from rpm to rad/s is
 2 π rad  1 min 
1 rpm  1 revolution /min 


 1 revolution  60 s 
π
 rad/s
30
66
Example 13.8 Analysis of Angular
Motion
a)The angular acceleration is
d

 0.02
dt
Separating the variables, we have
d

 0.02 dt
67
Example 13.8 Analysis of Angular
Motion
Defining t = 0 to be the time at which the fuel is
turned off:
1000 π 30 d
t
   0.02 dt

10,000 π 30

0
Evaluating the integrals and solving for t, we obtain,
1   10,000π 30 

t 
  115 s
 ln 
 0.02   1000π 30 
68
Example 13.8 Analysis of Angular
Motion
b)Writing the angular acceleration as
d d d d



  0.02
dt d dt d
Separating the variables, we have
d  0.02 d
69
Example 13.8 Analysis of Angular
Motion
Defining θ = 0 to be the angular position at which
the fuel is turned off:
1000 π 30

10,000 π 30 d  0  0.02 d
Solving for θ yields,
1 

 
10,000π 30  1000π 30
 0.02 
 15,000π rad  7500 rev
70