Chapter 13: Motion of a Point
Study of motion involves describing and analyzing
the motion of a point in space (either along an
arbitrary path or trajectory) from its position,
velocity and acceleration using various coordinate
systems.
1
Chapter Outline:
 Position, Velocity and Acceleration
 Straight-Line Motion
 Curvilinear Motion
 Relative Motion
 Chapter Summary
2
13.1 Position, Velocity and
Acceleration
A reference frame is a coordinate system
specifying the positions of points by specifying the
components of the position vector r relative to the
origin. Shown in Fig 13.1(a) and (b) are examples
for specifying positions of objects in a room and
an airplane.
3
13.1 Position, Velocity and
Acceleration
We can describe the position of a point P relative
to a given reference frame with origin O by the
position vector r from O to P as seen in Fig
13.2(a).
4
13.1 Position, Velocity and
Acceleration
Suppose P is in a motion relative to the chosen
reference frame, so the r is a function of time t
(see Fig 13.2(b)).
Thus, we can express this
by notation
r  r (t )
5
13.1 Position, Velocity and
Acceleration
The velocity of P relative to the given reference
frame at time t is defined by
dr
r (t  t )  r (t )
v   lim
dt t 0
t
(13.1)
where the vector r(t+Δt) – r(t) is the change in
position, or displacement, of P during the interval
of time Δt (refer to Fig 13.2(c)).
6
13.1 Position, Velocity and
Acceleration
Thus, the velocity v is the rate of change of
position of P.
7
13.1 Position, Velocity and
Acceleration
Time derivative of the sum of two vector functions
u and w is
d
du dw
u  w   
dt
dt dt
Time derivative of the product of a scalar function
f and a vector function u is
d  fu  df
du
 u f
dt
dt
dt
8
13.1 Position, Velocity and
Acceleration
The acceleration of P relative to the given
reference frame at time t is defined by
dv
v(t  t )  v(t )
a
 lim
dt t 0
t
(13.2)
where v(t+Δt) – v(t) is the change in velocity of P
during the interval of time Δt. (see Fig 13.3)
9
13.1 Position, Velocity and
Acceleration
Acceleration is the rate of change of the velocity of
P at time t i.e.
d  dr 
d 2r
2
or
(
m
/
s
)
 
2
dt  dt 
dt
10
13.1 Position, Velocity and
Acceleration
Let O’ be an arbitrary fixed point and r’ be the
position vector from O’ to P (refer to Fig 13.4(a)).
Velocity of P relative to O’
is
dr '
v' 
dt
Velocity of P relative to
origin O is
dr
v
dt
11
13.1 Position, Velocity and
Acceleration
To show v’ = v, let R
be the vector from O
to O’ (refer to Fig
13.4(b)), such that
r'  r  R
12
13.1 Position, Velocity and
Acceleration
Since the vector R is constant, the velocity of P
relative to O’ is
dr ' dr dR dr
v' 
 
 v
dt dt dt dt
and it is given that from definition,
dv'
dv
a' 
and a 
dt
dt
13
13.1 Position, Velocity and
Acceleration
Therefore, v’ = v and a’ = a.
Thus the velocity and acceleration of a point P
relative to a given reference frame do not depend
on the location of the fixed reference point used to
specify the position of P
14
13.2 Straight-Line Motion
 Description of the Motion
Consider a straight line through the origin O
of a given reference frame. Assuming the
direction of the line relative to the reference
frame is fixed, we can specify the position of
a point P relative to O by a coordinate s
measured along the line from O to P (Fig
13.5(a)).
15
13.2 Straight-Line Motion
Since s is defined to be positive to the right, thus
1. s is positive when P is to the right of O
2. s is negative when P is to the left of O
The displacement of P during an interval of time
from to to t is the change in the position s(t) –
s(to), where s(t) denotes the position at time t.
16
13.2 Straight-Line Motion
From Fig 13.5(b), by introducing a unit vector e
that is parallel to the line in positive s direction, the
position vector of P relative to O is
r  se
Because the magnitude and direction of e are
de
constant, such that
 0 , therefore the velocity
dt
of P relative to O is
dr ds
v  e
dt dt
17
13.2 Straight-Line Motion
Writing the velocity vector as v  ve , we can
obtain the scalar equation
ds
v
dt
where the velocity v of point P along the straight
line is the rate of change of position s.
18
13.2 Straight-Line Motion
From Fig 13.6, v is equal to the slope at time t of
the line tangent to the graph of s as a function of
time. Therefore the acceleration of P relative to O
is
dv d
dv
a
 ve   e
dt dt
dt
19
13.2 Straight-Line Motion
Writing the acceleration vector as a  ae , we
obtain the scalar equation
dv d 2 s
a
 2
dt dt
where acceleration a
is equal to the slope at
time t of the line
tangent to the graph of v
as a function of time.
(Fig 13.7)
20
13.2 Straight-Line Motion
By introducing the unit vector e, the position is
specified by the coordinate s and the velocity and
acceleration are governed by the equations
ds
v
dt
(13.3)
dv
a
dt
(13.4)
21
13.2 Straight-Line Motion
Applying the chain rule of differential calculus, the
derivation of velocity w.r.t time is
dv dv ds

dt ds dt
An alternative expression for the acceleration is
dv
a v
ds
(13.5)
22
13.2 Straight-Line Motion
 Analysis of the Motion
In some cases, the position s of a point of an
object is known as a function of time. Hence
methods such as radar and laser-Doppler
interferometry are used. In this case, we can
obtain the velocity and acceleration as
functions of time from Eqn (13.3) & (13.4) by
differentiation.
23
13.2 Straight-Line Motion
For example in Fig 13.8, the position of the truck
from t = 2 s to t = 4 s is given by the eqn
1 3
s  6 t m
3
Thus, the velocity and acceleration of the truck
during that interval of time are
ds 2
dv
v   t m/s and a   2t m/s 2
dt
dt
24
13.2 Straight-Line Motion
• Acceleration specified as a function of time
if the acceleration is a known function of time a(t),
we can integrate the relation
dv
 a(t )
(13.6)
dt
w.r.t time to determine the velocity as a function of
time such that
v   a(t ) dt  A
25
13.2 Straight-Line Motion
Given A is an integration constant, we can
integrate the relation
ds
v
dt
(13.7)
to determine the position as a function of time,
s   v dt  B
where B is another integration constant.
26
13.2 Straight-Line Motion
To determine the constants A and B, we can write
Eqn (13.6) as
dv  a (t ) dt
and integrate in terms of definite integrals:
v
t
v0 dv  t0 a(t ) dt
(13.8)
where v0 is the velocity at time t0
v is the velocity at an arbitrary time t
27
13.2 Straight-Line Motion
Evaluating the integral on the left side of Eqn
(13.8), we obtain an expression for the velocity as
a function of time:
t
v  v0   a(t ) dt
t0
(13.9)
Writing Eqn (13.7) as ds  v dt , and integrating it to
obtain:
s
t
s0 ds  t0 v dt
28
13.2 Straight-Line Motion
Once again, by evaluating the integral on the left
side, we obtain the position as a function of time:
t
s  s0   v(t ) dt
t0
(13.10)
It is recommended that straight-line motion
problems be solved by using Eqns (13.3)-(13.5).
It will be demonstrated in later examples.
29
13.2 Straight-Line Motion
From Eqns (13.9) & (13.10), we can observed that
•The area defined by the acceleration graph of P
from t0 to t is equal to the change in velocity from t0
to t. (Fig 13.9(a))
30
13.2 Straight-Line Motion
• The area defined by the velocity graph of P
from t0 to t is equal to the change in position
from t0 to t. (Fig 13.9(b))
31
13.2 Straight-Line Motion
• Constant Acceleration
Let the acceleration be a known constant a0. From
Eqns (13.9) & (13.10), the velocity and position as
functions of time are
v  v0  a0 (t  t0 )
(13.11)
1
s  s0  v0 (t  t0 )  a0 (t  t0 )2
2
(13.12)
32
13.2 Straight-Line Motion
Notice that s0 and v0 are the position and velocity
at time t0. If the acceleration is constant, the
velocity is a linear function of time. From Eqn
(13.5), we can write the acceleration as
dv
a0  v
ds
Rewriting as v dv  a0 ds , and integrating,
v
s
v0 v dv  s0 a0 ds
33
13.2 Straight-Line Motion
We can obtain an equation for the velocity as a
function of position:
v 2  v02  2a0 ( s  s0 )
(13.13)
Eqns (13.11) to (13.13) are suitable to be used
only when the acceleration is constant.
34
Example 13.1 Straight-Line Motion
with Constant Acceleration
 Question
Engineers testing a vehicle that will be
dropped by parachute estimate that the
vertical velocity of the vehicle when it reaches
the ground will be 6 m/s. If they drop the
vehicle from the test
rig in Fig 13.10, from
what height h should
they drop it to match the
impact velocity of the
parachute drop?
35
Example 13.1 Straight-Line Motion
with Constant Acceleration
 Strategy
If only the significant force acting on an object
near the earth’s surface is its weight, the
acceleration of the object is approximately
constant and equal to the acceleration due to
gravity at sea level. Therefore we can assume
that the vehicle’s acceleration during its short fall
is g = 9.81 m/s2. Integrating Eqns (13.3) & (13.4)
to obtain the vehicle’s velocity and position, and
hence its position when its velocity is 6 m/s.
36
Example 13.1 Straight-Line Motion
with Constant Acceleration
 Solution
let t = 0 when the vehicle is dropped, and let s
be the position of the bottom of the cushioning
material beneath the vehicle relative to its
position at t = 0 (Fig a). The vehicle’s
acceleration is a = 9.81 m/s2.
From Eqn (13.4),
dv
 a  9.81 m/s 2
dt
37
Example 13.1 Straight-Line Motion
with Constant Acceleration
Integrating, we obtain
v  9.81 t  A
where A is an integration constant.
Because the vehicle is at rest when it is released,
v = 0 at t = 0. Therefore A = 0 and hence,
v  9.81 t m/s
38
Example 13.1 Straight-Line Motion
with Constant Acceleration
Substitute the result into Eqn (13.3), we have
ds
 v  9.81 t
dt
and integrating it to obtain,
s  4.91 t 2  B
When t = 0, s = 0 and B = 0, hence
s  4.91 t
2
39
Example 13.1 Straight-Line Motion
with Constant Acceleration
The time necessary for the vehicle to reach 6 m/s
as it falls is
v
6 m/s
t

 0.612 s
9.81 m/s 9.81 m/s
Hence, the required height h is
h  4.91 t  4.91(0.612)  1.83 m
2
2
40
Example 13.2 Graphical Solution of
Straight-Line Motion
 Question
The cheetah can run as fast as 120 km/h. If
assuming the animal’s acceleration is constant
and that it reaches top speed in 4 s, what
distance can the cheetah cover in 10 s?
41
Example 13.2 Graphical Solution of
Straight-Line Motion
 Strategy
the acceleration has a constant value for the
first 4 s and is then zero. We can determine
the distance traveled during each of these
“phases” of the motion and sum them to
obtain the total distance covered, either
analytically or graphically.
42
Example 13.2 Graphical Solution of
Straight-Line Motion
 Solution
the top speed in terms of feet per second is
120  1000 m
1.6 km/h 
 33.33 m/s
3600 s
1. Analytical Method
let a0 be the acceleration during the first 4 s.
we integrate Eqn (13.4) to get
43
Example 13.2 Graphical Solution of
Straight-Line Motion
v
t
0 dv  0 a0 dt
vv0  a0 t t0
v  0  a0 (t  0)
obtaining the velocity as a function of time during
the first 4 s:
v  a0 t m/s
44
Example 13.2 Graphical Solution of
Straight-Line Motion
When t = 4 s, v = 33.33 m/s and a0 = 8.33 m/s2.
since the velocity during the first 4 s is v = 8.33t
m/s, integrating Eqn (13.3) we have
s
t
0 ds  0 8.33t dt
s 
s
0
2 t
t
 8.33 
 2 0
t2
s  0  8.33(  0)
2
45
Example 13.2 Graphical Solution of
Straight-Line Motion
Obtaining the position as a function of time during
the first 4 s:
s  4.17 t 2 m
At t = 4 s, the position is s = 4.17(4)2 = 66.7 m.
From t = 4 s to t = 10 s, v = 33.33 m/s. Rewriting
Eqn (13.3) as
ds  v dt  33.33 dt
46
Example 13.2 Graphical Solution of
Straight-Line Motion
Integrate to determine the distance traveled during
the second phase of the motion,
s
10
0 ds  4
s 
s
0
33.33 dt
 33.33t 
10
4
s  0  33.33(10  4)
s  200 m
Hence the total distance traveled in 10 s is 66.7 m
+ 200 m = 266.7 m or 0.267 km.
47
Example 13.2 Graphical Solution of
Straight-Line Motion
2. Graphical method
By drawing a graph of the cheetah’s velocity
as a function of time in Fig (a), the total
distance covered is the sum of the areas
during the two phases of motion. Acceleration
is constant during the first 4 s of motion, so the
graph is linear from v = 0 at t = 0 to v = 33.33
m/s at t = 4 s. Velocity is constant during the
last 6 s.
48
Example 13.2 Graphical Solution of
Straight-Line Motion
Total distance covered by the cheetah is
1
4 s 33.33 m/s   6 s 33.33 m/s   66.7 m  200 m
2
 266.7 m
49
Example 13.3 Acceleration that is a
Function of Time
 Question
Suppose that the acceleration of the train in Fig.
13.12 during the interval of time from t = 2 s to t
= 4 s is a = 2t m/s2, and at t = 2 s its velocity is v
= 180 km/h. what is the train’s velocity at t = 4 s,
and what is its displacement (change in position)
from t = 2 s to t = 4 s?
50
Example 13.3 Acceleration that is a
Function of Time
 Strategy
We can integrate Eqns (13.3) & (13.4) to
determine the train’s velocity and position as
functions of time.
51
Example 13.3 Acceleration that is a
Function of Time
 Solution
The velocity at t = 2 s is
 1000 m  1 h 
180 km/h 

  50 m/s
 1 km  3600 s 
Rewrite Eqn (13.4) as
dv  a dt  2t dt
52
Example 13.3 Acceleration that is a
Function of Time
Given the condition v = 50 m/s at t = 2 s,
v
t
50 dv  2 2 t dt
v
v
50


2t
t 2
2
2
v  50  t  2
v  t 2  46 m/s
53
Example 13.3 Acceleration that is a
Function of Time
Rewriting Eqn (13.3), we have


ds  v dt  t  46 dt
2
Integrating at t = 2 s, s = 0:
s
0
t
ds   (t 2  46) dt
2
t
t

s     46 t 
3
2
s
0
3
54
Example 13.3 Acceleration that is a
Function of Time
Hence yielding the position as a function of time,
t3
23
s  0   46 t   46(2)
3
3
t3
s   46 t  94.7 m
3
55
Example 13.3 Acceleration that is a
Function of Time
Using our equations for the velocity and position,
the velocity at t = 4 s is
v  42  46  62 m/s
Therefore, the displacement from t = 2 s to t = 4 s
is
1 43  46(4)  94.7   0  111 m
3

56