t t+Δt

•Motion along a straight line

Position at any instance of time t

:: specified by its distance s measured from some convenient reference point O
fixed on the line

:: (disp. is negative if the particle moves in the negative s-direction).

Velocity of the particle: Acceleration of the particle:

ds dv d 2s
v  s Both are a  v or a  2  s
dt vector quantities dt dt
+ve or –ve depending +ve or –ve depending on whether
on +ve or –ve displacement velocity is increasing or decreasing
vdv  a ds or s ds  s ds
Rectilinear Motion:
Graphical Interpretations
Using s-t curve, v-t & a-t curves can be plotted.
Area under v-t curve during time dt = vdt == ds
• Net disp from t1 to t2 = corresponding area under
v-t curve  s t
s ds   vdt
2 2

1 t1

or s2 - s1 = (area under v-t curve)

Area under a-t curve during time dt = adt == dv

• Net change in vel from t1 to t2 = corresponding
area under a-t curve  v t
v dv   adt
2 2

1 t1

or v2 - v1 = (area under a-t curve)

Rectilinear Motion:
Graphical Interpretations
Two additional graphical relations:
Area under a-s curve during disp ds= ads == vdv
• Net area under a-s curve betn position
coordinates s1 and s2  v2 s2
v
1
vdv  ads
s1

or ½ (v22 – v12) = (area under a-s curve)

Slope of v-s curve at any point A = dv/ds
• Construct a normal AB to the curve at A. From
similar triangles:
CB dv dv
  CB  v  a (accelerati on)
v ds ds
• Vel and posn coordinate axes should have the same
numerical scales so that the accln read on the x-axis in
2
Analytical Integration to find the position coordinate
Acceleration may be specified as a function of time, velocity, or position
coordinate, or as a combined function of these.

(a) Constant Acceleration

At the beginning of the interval  t = 0, s = s0, v = v0
For a time interval t: integrating the following two equations
dv
a
dt
vdv  a ds
Substituting in the following equation and integrating will give the position
coordinate:
ds
v
dt
Equations applicable for Constant Acceleration and for time interval 0 to t
Analytical Integration to find the position coordinate
(b) Acceleration given as a function of time, a = f(t)

At the beginning of the interval  t = 0, s = s0, v = v0

For a time interval t: integrating the following equation
dv  dv
a f (t ) 
dt dt
Substituting in the following equation and integrating will give the position
coordinate:
ds
v
dt
Alternatively, following second order differential equation may be solved to
get the position coordinate:
d 2s
a  2  s  s  f (t )
dt
Analytical Integration to find the position coordinate
(c) Acceleration given as a function of velocity, a = f(v)

At the beginning of the interval  t = 0, s = s0, v = v0

For a time interval t: Substituting a and integrating the following equation
dv  dv
a f (v ) 
dt dt
Solve for v as a function of t and integrate the following equation to get the
position coordinate: ds
v
dt
Alternatively, substitute a = f(v) in the following equation and integrate to
get the position coordinate :

vdv  a ds
Analytical Integration to find the position coordinate
(d) Acceleration given as a function of displacement, a = f(s)

At the beginning of the interval  t = 0, s = s0, v = v0

For a time interval t: substituting a and integrating the following equation
vdv  a ds

Solve for v as a function of s : v = g(s), substitute in the following equation

and integrate to get the position coordinate:
ds
v
dt

It gives t as a function of s. Rearrange to obtain s as a function of t to get

the position coordinate.

In all these cases, if integration is difficult, graphical, analytical, or

computer methods can be utilized.
Example
Position coordinate of a particle confined to move along a straight line is given by
s = 2t3 – 24t + 6, where s is measured in meters from a convenient origin and t is
in seconds. Determine: (a) time reqd for the particle to reach a velocity of 72 m/s
from its initial condition at t = 0, (b) acceleration of the particle when v = 30 m/s,
and (c) net disp of the particle during the interval from t = 1 s to t = 4 s.
Solution

Differentiating s = 2t3 – 24t + 6  v = 6t2 – 24 m/s

 a = 12t m/s2
(a) v = 72 m/s  t = ± 4 s
(- 4 s happened before initiation of motion  no physical interest.)
t=4s

(b) v = 30 m/s  t = 3 sec  a = 36 m/s2

(c) t = 1 s to 4 s. Using s = 2t3 – 24t + 6

Δs = s4 – s1 = [2(43) – 24(4) +6] – [2(13) – 24(1) + 6]
Δs = 54 m
Plane Curvilinear Motion:
Between A and A’:
Average velocity of the particle : vav = Δr/ Δt
 A vector whose direction is that of Δr and whose
magnitude is magnitude of Δr/ Δt
Average speed of the particle = Δs/ Δt
Instantaneous velocity of the particle is defined as
the limiting value of the average velocity as the time
interval approaches zero 

 v is always a vector tangent to the path

Extending the definition of derivative of a scalar to include vector quantity:

Derivative of a vector is a vector having a magnitude and a direction.

Magnitude of v is equal to speed (scalar)

Plane Curvilinear Motion
Magnitude of the derivative:
dr / dt  r  s  v  v
 Magnitude of the velocity or the speed

Derivative of the magnitude:

d r / dt  dr / dt  r
 Rate at which the length of the position vector is changing

Velocity of the particle at A  tangent vector v

Velocity of the particle at A’  tangent vector v’
 v’ – v = Δv
 Δv Depends on both the change in magnitude of v and
on the change in direction of v.
Plane Curvilinear Motion
Between A and A’:
Average acceleration of the particle : aav = Δv/ Δt
 A vector whose direction is that of Δv and whose
magnitude is the magnitude of Δv/ Δt
Instantaneous accln of the particle is defined as
the limiting value of the average accln as the time
interval approaches zero 

By definition of the derivative:

 In general, direction of the acceleration of a particle

in curvilinear motion neither tangent to the path
nor normal to the path.
 Acceleration component normal to the path points
toward the center of curvature of the path.
 NORMAL AND TANGENTIAL COMPONENTS
(Section 12.7)
When a particle moves along a curved path, it is sometimes convenient
to describe its motion using coordinates other than Cartesian. When the
path of motion is known, normal (n) and tangential (t) coordinates are
often used.

In the n-t coordinate system, the

origin is located on the particle
(the origin moves with the
particle).

The t-axis is tangent to the path (curve) at the instant considered,

positive in the direction of the particle’s motion.
The n-axis is perpendicular to the t-axis with the positive direction
toward the center of curvature of the curve.
 NORMAL AND TANGENTIAL COMPONENTS
(continued)
The positive n and t directions are
defined by the unit vectors un and ut,
respectively.

The center of curvature, O’, always

lies on the concave side of the curve.
The radius of curvature, r, is defined
as the perpendicular distance from
the curve to the center of curvature at
that point.

The position of the particle at any instant is defined by the

distance, s, along the curve from a fixed reference point.
 VELOCITY IN THE n-t COORDINATE SYSTEM

The velocity vector is always

tangent to the path of motion
(t-direction).

The magnitude is determined by taking the time derivative of

the path function, s(t). .
v = vut where v = s = ds/dt
Here v defines the magnitude of the velocity (speed) and
ut defines the direction of the velocity vector.
ACCELERATION IN THE n-t COORDINATE SYSTEM
Acceleration is the time rate of change
. of velocity:
.
a = dv/dt = d(vut)/dt = vut + vut
.
Here v represents the change in
.
the magnitude of velocity and ut
represents the rate of change in
the direction of ut.

After mathematical manipulation,

the acceleration vector can be
expressed as:
.
a = vut + (v2/r)un = atut + anun.
ACCELERATION IN THE n-t COORDINATE SYSTEM
(continued)
There are two components to the
acceleration vector:
a = at ut + an un

• The tangential component is tangent to the curve and in the

direction of increasing or decreasing velocity.
.
at = v or at ds = v dv
• The normal or centripetal component is always directed
toward the center of curvature of the curve. an = v2/r
• The magnitude of the acceleration vector is
a = [(at)2 + (an)2]0.5
 SPECIAL CASES OF MOTION
There are some special cases of motion to consider.

1) The particle moves along a straight line. .

r  => an = v /r = 0 => a = at = v
2

The tangential component represents the time rate of change in

the magnitude of the velocity.

2) The particle moves along a curve at constant speed.

.
at = v = 0 => a = an = v2/r

The normal component represents the time rate of change in the

direction of the velocity.
 SPECIAL CASES OF MOTION
(continued)
3) The tangential component of acceleration is constant, at = (at)c.
In this case,
s = so + vot + (1/2)(at)ct2
v = vo + (at)ct
v2 = (vo)2 + 2(at)c(s – so)
As before, so and vo are the initial position and velocity of the
particle at t = 0. How are these equations related to projectile
motion equations? Why?
4) The particle moves along a path expressed as y = f(x).
The radius of curvature, r, at any point on the path can be
calculated from
[ 1 + (dy/dx)2 ]3/2
r = ________________
d2y/dx 2
 THREE-DIMENSIONAL MOTION

If a particle moves along a space

curve, the n and t axes are defined as
before. At any point, the t-axis is
tangent to the path and the n-axis
points toward the center of curvature.
The plane containing the n and t axes
is called the osculating plane.

A third axis can be defined, called the binomial axis, b. The

binomial unit vector, ub, is directed perpendicular to the osculating
plane, and its sense is defined by the cross product ub = ut x un.
There is no motion, thus no velocity or acceleration, in the
binomial direction.
 EXAMPLE PROBLEM
Given: Starting from rest, a motorboat
travels around a circular path of
r = 50 m at a speed that
increases with time,
v = (0.2 t2) m/s.
Find: The magnitudes of the boat’s
velocity and acceleration at
the instant t = 3 s.
Plan: The boat starts from rest (v = 0 when t = 0).
1) Calculate the velocity at t = 3s using v(t).
2) Calculate the tangential and normal components of
acceleration and then the magnitude of the
acceleration vector.
 EXAMPLE
(continued)
Solution:
1) The velocity vector is v = v ut , where the magnitude is
given by v = (0.2t2) m/s. At t = 3s:
v = 0.2t2 = 0.2(3)2 = 1.8 m/s
.
2) The acceleration vector is a = atut + anun = vut + (v2/r)un.
.
Tangential component: at = v = d(.2t2)/dt = 0.4t m/s2
At t = 3s: at = 0.4t = 0.4(3) = 1.2 m/s2

Normal component: an = v2/r = (0.2t2)2/(r) m/s2

At t = 3s: an = [(0.2)(32)]2/(50) = 0.0648 m/s2

The magnitude of the acceleration is

a = [(at)2 + (an)2]0.5 = [(1.2)2 + (0.0648)2]0.5 = 1.20 m/s2
 Summary

v = vut v = s = ds/dt

a = vut + (v2/r)un = atut + anun.

at = v or at ds = v dv
2 3 2
𝑑𝑦
1+ 𝑑𝑥
𝜌=
𝑑2 𝑦
𝑑𝑥 2

𝑑𝑦
𝑡𝑎𝑛𝜃 =
𝑑𝑥
 Summary

Special Case: constant acceleration

1D

∆s = v0tt + 1/2att2 ∆x = vx0t + 1/2axt2

∆s = ½(v0t + vt)t ∆x = ½(vx0 + vx)t
vt = v0t + att vx = vx0 + axt
2at∆s = (vt)2  (vt)2 2ax∆x = (vx)2  (vx0)2
 Summary
2 3 2
𝑑𝑦
1+ 𝑑𝑥
𝜌=
𝑑2 𝑦
𝑑𝑥 2
𝑑𝑦 𝑢𝑡 = 𝑖𝑐𝑜𝑠𝜃 + 𝑗𝑠𝑖𝑛𝜃
Have 𝑦(𝑥) 𝑡𝑎𝑛𝜃 =
𝑢𝑛 = −𝑖𝑠𝑖𝑛𝜃 + 𝑗𝑐𝑜𝑠𝜃
𝑑𝑥
𝑥 2
𝑑𝑦
𝑠(𝑥) = 1+ 𝑑𝑥
0 𝑑𝑥
s(𝑡)
𝑑𝑥 = 𝑣𝑥 𝑑𝑡 𝑥(𝑡)
Have 𝑣𝑥 (𝑡) 𝑑𝑣𝑥
𝑎𝑥 =
𝑑𝑡

𝑑𝑠 = 𝑣𝑡 𝑑𝑡 𝑠(𝑡)
Have 𝑣𝑡 (𝑡) 𝑑𝑣𝑡
𝑎𝑛 =
𝑣𝑡2
𝑎𝑡 =
𝑑𝑡 𝜌
 Summary

Have 𝑎𝑡 (𝑡) 𝑑𝑣𝑡 = 𝑎𝑡 𝑑𝑡 𝑣𝑡 (𝑡) 𝑠(𝑡)

Have 𝑎𝑡 (𝑠) 𝑎𝑡 𝑑𝑠 = 𝑣𝑡 𝑑𝑣𝑡 𝑣𝑡 (𝑠)

In many kinematics problems, the motion of one object will
depend on the motion of another object.
The blocks in this figure are
connected by an inextensible cord
wrapped around a pulley. If block
A moves downward along the
inclined plane, block B will move
up the other incline.

The motion of each block can be related mathematically by

defining position coordinates, sA and sB. Each coordinate axis is
defined from a fixed point or datum line, measured positive along
each plane in the direction of motion of each block.
 In this example, position
coordinates sA and sB can be
defined from fixed datum lines
extending from the center of
the pulley along each incline
to blocks A and B.

If the cord has a fixed length, the position coordinates sA

and sB are related mathematically by the equation
sA + lCD + sB = lT

Here lT is the total cord length and lCD is the length of cord
passing over arc CD on the pulley.
 The velocities of blocks A and B
can be related by differentiating
the position equation. Note that
lCD and lT remain constant, so
dlCD/dt = dlT/dt = 0

dsA/dt + dsB/dt = 0 => vB = -vA

The negative sign indicates that as A moves down the incline
(positive sA direction), B moves up the incline (negative sB
direction).
Accelerations can be found by differentiating the velocity
expression. Prove to yourself that aB = -aA .
 Consider a more complicated
example. Position coordinates (sA
and sB) are defined from fixed
datum lines, measured along the
direction of motion of each block.

Note that sB is only defined to the

center of the pulley above block
B, since this block moves with the
pulley. Also, h is a constant.

The red colored segments of the cord remain constant in length

during motion of the blocks.
 The position coordinates are related
by the equation
2sB + h + sA = l
Where l is the total cord length minus
the lengths of the red segments.
Since l and h remain constant during
the motion, the velocities and
accelerations can be related by two
successive time derivatives:
2vB = -vA and 2aB = -aA

When block B moves downward (+sB), block A moves to the

left (-sA). Remember to be consistent with the sign convention!
 This example can also be worked
by defining the position coordinate
for B (sB) from the bottom pulley
instead of the top pulley.

The position, velocity, and

acceleration relations then become
2(h – sB) + h + sA = l
and 2vB = vA 2aB = aA

Prove to yourself that the results are the same, even if the sign
conventions are different than the previous formulation.
particles moving along rectilinear paths (only the magnitudes of
velocity and acceleration change, not their line of direction).
1. Define position coordinates from fixed datum lines,
along the path of each particle. Different datum lines can
be used for each particle.
2. Relate the position coordinates to the cord length.
Segments of cord that do not change in length during the
motion may be left out.
3. If a system contains more than one cord, relate the
position of a point on one cord to a point on another
cord. Separate equations are written for each cord.
4. Differentiate the position coordinate equation(s) to relate
velocities and accelerations. Keep track of signs!
 RELATIVE POSITION
(Section 12.10)
The absolute position of two
particles A and B with respect to
the fixed x, y, z reference frame are
given by rA and rB. The position of
B relative to A is represented by
rB/A = rB – rA

Therefore, if rB = (10 i + 2 j ) m
and rA = (4 i + 5 j ) m,
then rB/A = (6 i – 3 j ) m.
 RELATIVE VELOCITY
To determine the relative velocity of B
with respect to A, the time derivative of
the relative position equation is taken.
vB/A = vB – vA
or
vB = vA + vB/A

In these equations, vB and vA are called absolute velocities

and vB/A is the relative velocity of B with respect to A.

Note that vB/A = - vA/B .

RELATIVE ACCELERATION

The time derivative of the relative

velocity equation yields a similar
vector relationship between the
absolute and relative accelerations of
particles A and B.
aB/A = aB – aA
or
aB = aA + aB/A
 SOLVING PROBLEMS

Since the relative motion equations are vector equations,

problems involving them may be solved in one of two ways.
For instance, the velocity vectors in vB = vA + vB/A could be
written as Cartesian vectors and the resulting scalar
equations solved for up to two unknowns.

Alternatively, vector problems can be solved “graphically” by

use of trigonometry. This approach usually makes use of the
law of sines or the law of cosines.

Could a CAD system be used to solve these types of

problems?
 LAWS OF SINES AND COSINES

Since vector addition or subtraction forms

C
a triangle, sine and cosine laws can be
a b
applied to solve for relative or absolute
velocities and accelerations. As review,
A
B their formulations are provided below.
c
Law of Sines: a b c
= =
sin A sin B sin C

Law of Cosines: a 2 = b 2 + c 2 - 2 bc cos A

b = a + c - 2 ac cos B
2 2 2

2 2 2
c = a + b - 2 ab cos C
 EXAMPLE

Given: vA = 600 km/hr

vB = 700 km/hr
Find: vB/A

Plan:
a) Vector Method: Write vectors vA and vB in Cartesian
form, then determine vB – vA

b) Graphical Method: Draw vectors vA and vB from a

common point. Apply the laws of sines and cosines to
determine vB/A.
 EXAMPLE
(continued)

Solution:
a) Vector Method:

vA = 600 cos 35 i – 600 sin 35 j

= (491.5 i – 344.1 j ) km/hr
vB = -700 i km/hr

vB/A = vB – vA = (- 1191.5 i + 344.1 j ) km/hr

vB /A = 1191. 5
( )2
+ ( 344.1 ) 2
= 1240. 2 km
hr
where q = tan
- 1 344 .1
( ) = 16.1 q
1191 . 5
 EXAMPLE
(continued)
b) Graphical Method: vB = 700 km/hr
q
Note that the vector that measures 145 
the tip of B relative to A is vB/A.

Law of Cosines:
vB/A2 = ( 700 ) 2 + ( 600 ) - 2 ( 700)(600 )cos 145 
2

vB/A = 1240 . 2 km
hr
Law of Sines:
vB/A vA
= or q = 16 . 1 
sin(145° ) sin q