Chapter 2: Motion

Part 1:
Motion Along a Straight Line
Part 2:
Motion in Two or Three Dimensions

Prof. Wei BAO

AC1-G5106
3442-7848

1
Chapter 1 summary

 Basic units and dimensions

─ Dimensional analysis
 Vectors and scalars
─ Vector addition and subtraction
─ Vector components: unit vectors
─ Vector multiplication: scalar product and vector
product
 Calculus
─ Differentiation of simple functions
─ Partial differentiation

2
Topics for Chapter 2 part 1

 Straight-line motion in terms of velocity and

acceleration
 Average and instantaneous velocity and
average and instantaneous acceleration
 Straight-line motion with constant and varying
acceleration
 Graphical representation of position, velocity
and acceleration for straight-line motion

3
What is Motion?
 Motion is a change in position of an object with respect to time.
 Motion is typically described in terms of displacement, distance,
velocity, acceleration, time and speed.
─ Change of the basket ball position as a function of time
─ Change of missile and train position with time

4
Why do we need to study motion/mechanics?

 Design of engine
 Design of locomotives (car,
train, airplane, rocket,
missile)
 Understand human motion in
sport
 Design robots, artificial limbs
 Understand movement of
electrons in solid-electrical
conduction

5
Introduction
 Kinetics is the study of motion (position as a function
of time).
 In 1 dimension, motion is the change of x coordinate
as a function of time t. In 3 dimension, it is the change
of position vector 𝑟𝑟 as a function of time.
 The objective is to find the position as a function of
time
 Velocity and acceleration are physical quantities
used to describe motion.
 If you know the velocity and acceleration, then you can
find the position (coordinate) as function of time.
 Classical mechanics is fundamentally based on
Newton’s Laws of Motion

6
Motions

7
Displacement, time, and average velocity
 A particle moving along the x-axis has a
coordinate x. x is a function of time t. So it
is in motion.
 The change in the particle’s coordinate is
Δx = x2 - x1, in time Δt.
─ Δx is the displacement.
 The average x-velocity of the particle is
∆x
vav−x = . Displacement divided by time.
∆t
It is the average rate of change of
displacement.
 If you know average velocity vav−x you can
find displacement ∆x = vav−x ∆t and so the
position (motion).
 Average velocity is a vector quantity and A position-time graph (an x-
can be positive or negative t graph) shows the
particle’s position x as a
 Speed is the magnitude of velocity ─scalar function of time t.

8
Concepts of velocity and acceleration

 Motion is a change of position with time

 We are interested in how fast the position
changes with time, so we have the concept of
velocity and speed.
 We are interested in how fast the motion is
changed with time, i.e. how fast the velocity is
changing with time, so we have the concept of
acceleration.
 Velocity and acceleration are quantities used to
describe motion

9
10
Instantaneous velocity
 The instantaneous velocity is the
velocity at a specific instant of time or
specific point along the path and is
given by vx = dx/dt. (differentiation of x
with respect to t)
 It is obtained from the average velocity
by making the time interval smaller and
smaller. When the interval is very very
small, it becomes the velocity of that
moment of time t.
∆x dx
vx = lim =
∆t→0 ∆t dt

Uniform motion, Constant velocity:

Uniform motion means the instantaneous velocity is a
constant. So, the average velocity equals the instantaneous
velocity and is also a constant.

11
Finding instantaneous velocity on an x-t graph
At any point on an x-t graph, the instantaneous x-velocity is
equal to the slope of the tangent to the curve at that point.
∆x dx
vx = lim =
∆t→0 ∆t dt
 The average velocity from point A
∆𝑥𝑥 40 𝑚𝑚
to D is = = 13.33 m/s
∆𝑡𝑡 3 𝑠𝑠
 The instantaneous velocity at
point C is the tangent line at C
(blue line)
 The instantaneous velocity at
point D is negative (green)
 At what point is the instantaneous
velocity=0?

12
 Average acceleration
 Acceleration is the rate of change of velocity with time
 Average x-acceleration is acceleration over an interval of time
∆vx
∆𝑡𝑡, so that aav−x = (unit is meter/sec2)
∆t
 The instantaneous acceleration is the acceleration when ∆𝑡𝑡
∆vx dvx d dx d2 x
approaches 0, ax = lim = = =
∆t→0 ∆t dt dt dt dt2

Slope=instantaneous
You can find the change in 𝑑𝑑𝑣𝑣
acceleration 𝑥𝑥
velocity from the average 𝑑𝑑𝑑𝑑
acceleration ∆𝑣𝑣𝑥𝑥 = 𝑎𝑎𝑎𝑎𝑎𝑎−𝑥𝑥 ∆𝑡𝑡

Velocity, 𝑣𝑣𝑥𝑥
Uniform acceleration, constant Δv
acceleration means the ∆vx
instantaneous and average Δt aav−x =
∆t
acceleration are equal and is a
constant.
time
13
Motion with constant acceleration
For a particle with constant acceleration, the velocity changes at the same
rate throughout the motion, i.e. a linear curve in a v-t graph
𝑣𝑣𝑥𝑥

velocity, v
𝑎𝑎𝑥𝑥 𝑡𝑡

𝑣𝑣𝑜𝑜𝑜𝑜 vx = vox + ax t

𝑣𝑣𝑜𝑜𝑜𝑜

t time, t
Total area under the curve =total
change in position, i.e. displacement
1
𝑥𝑥 = 𝑥𝑥𝑜𝑜 + 𝑣𝑣𝑜𝑜𝑜𝑜 𝑡𝑡 + 𝑎𝑎𝑥𝑥 𝑡𝑡 2
2

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Area under the 𝐯𝐯𝐱𝐱 − 𝐭𝐭 curve gives the displacement
 ∆x = vx ∆t, is the area of a slice under the curve and also
the displacement in time Δt
 Adding the area of all the slices give you the total
displacement, which is the area under the curve

∆𝑥𝑥 = 𝑣𝑣𝑥𝑥 ∆𝑡𝑡

Shaded area is 𝑥𝑥 = ∑ 𝑣𝑣𝑥𝑥 ∆𝑡𝑡

15
What is the use of velocity and acceleration?
 In this course most of the problems are either constant
acceleration or constant velocity.
 However, you still need to know (with some
understanding) the concepts of instantaneous velocity
and acceleration, the concept of differentiation. (know
the definition !).
 Given the initial velocity (vox) at t=0 and the constant
acceleration (ax), we can find the velocity vx and position
x at time t.
 There are four equations (given below) which can be
used to do this job. You should be able to solve most of
the problems using these four equations

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Equations of Motion with Constant Acceleration
The four equations can be applied to any straight-line
motion with constant acceleration ax.

 The 1st equation gives the vx = vox + ax t

velocity at time t.
 The 2nd equation gives the 1 2
position of the particle at time t.
x = xo + vox t + ax t
2
 The 3rd equation gives the
velocity at t in terms of the vx2 = vox
2
+ 2ax x − xo
initial velocity, acceleration and
the distance travelled. vox + vx
x − xo = t
 The 4th equation gives the 2
distance travelled in terms of
the initial velocity, final velocity You should be able to solve most of
and time spent. the problems concerning linear motion
with constant acceleration using these
four equations

17
 Derivation of the Four Equations
vx = vox + ax t (1) Equation 1:
1 From the graph we see that
x = xo + vox t + ax t 2 (2)
2 ax t = vx − v0x , so
vx2 = vox
2
+ 2ax x − xo (3) vx = v0x + ax t (1)
vox +vx
x − xo = t (4) Equation 2 and 4:
2
From the graph:
1
𝑣𝑣𝑥𝑥 Green area= ax t 2 and blue area=vox t
2
Total area is the displacement ∶
velocity, v

1
x − xo = vox t + ax t 2
𝑎𝑎𝑥𝑥 𝑡𝑡 2
1
𝑣𝑣𝑥𝑥 x = xo + vox t + ax t 2 (2)
𝑣𝑣𝑜𝑜𝑜𝑜 2
Using equation 1, ax t = vx − vox
𝑣𝑣𝑜𝑜𝑜𝑜 So Equation 2 becomes
1
x − xo = vox t + vx − vox t
t time, t 2
v +v
Total area under the curve =total x − xo = ox x t (4)
2
change in position, i.e. displacement
18
 Derivation of the Four Equations
𝑣𝑣𝑥𝑥 = 𝑣𝑣𝑜𝑜𝑜𝑜 + 𝑎𝑎𝑥𝑥 𝑡𝑡 (1)
Equation 3:
1
𝑥𝑥 = 𝑥𝑥𝑜𝑜 + 𝑣𝑣𝑜𝑜𝑜𝑜 𝑡𝑡 + 𝑎𝑎𝑥𝑥 𝑡𝑡 2 (2) From Equation 1 and 2
2
𝑣𝑣𝑥𝑥 = 𝑣𝑣0𝑥𝑥 + 𝑎𝑎𝑥𝑥 𝑡𝑡 ⇒ 𝑣𝑣𝑜𝑜𝑥𝑥 = 𝑣𝑣𝑥𝑥 − 𝑎𝑎𝑥𝑥 𝑡𝑡
𝑣𝑣𝑥𝑥2 = 𝑣𝑣𝑜𝑜𝑜𝑜
2
+ 2𝑎𝑎𝑥𝑥 𝑥𝑥 − 𝑥𝑥𝑜𝑜 (3) 1
𝑥𝑥 − 𝑥𝑥𝑜𝑜 = 𝑣𝑣𝑜𝑜𝑜𝑜 𝑡𝑡 + 𝑎𝑎𝑥𝑥 𝑡𝑡 2
𝑥𝑥 − 𝑥𝑥𝑜𝑜 =
𝑣𝑣𝑜𝑜𝑜𝑜 +𝑣𝑣𝑥𝑥
𝑡𝑡 (4) 2
2 1
⇒ 𝑥𝑥 − 𝑥𝑥𝑜𝑜 = (𝑣𝑣𝑥𝑥 −𝑎𝑎𝑥𝑥 𝑡𝑡)𝑡𝑡 + 𝑎𝑎𝑥𝑥 𝑡𝑡 2
𝑣𝑣𝑥𝑥 2
Multiply both sides by 2𝑎𝑎𝑥𝑥 :
velocity, v

2𝑎𝑎𝑥𝑥 𝑥𝑥 − 𝑥𝑥𝑜𝑜 = 2𝑎𝑎𝑥𝑥 𝑣𝑣𝑥𝑥 𝑡𝑡 − 𝑎𝑎𝑥𝑥2 𝑡𝑡 2

𝑎𝑎𝑥𝑥 𝑡𝑡 From Equation 1, we know
𝑣𝑣𝑜𝑜𝑜𝑜 𝑣𝑣𝑥𝑥 𝑎𝑎𝑥𝑥 𝑡𝑡 = 𝑣𝑣𝑥𝑥 − 𝑣𝑣𝑜𝑜𝑜𝑜 , so
2𝑎𝑎𝑥𝑥 𝑥𝑥 − 𝑥𝑥𝑜𝑜
𝑣𝑣𝑜𝑜𝑜𝑜 = 2𝑣𝑣𝑥𝑥 (𝑣𝑣𝑥𝑥 − 𝑣𝑣𝑜𝑜𝑜𝑜 ) − 𝑣𝑣𝑥𝑥 − 𝑣𝑣𝑜𝑜𝑜𝑜 2

⇒ 2𝑎𝑎𝑥𝑥 𝑥𝑥 − 𝑥𝑥𝑜𝑜 = 𝑣𝑣𝑥𝑥2 − 𝑣𝑣𝑜𝑜𝑜𝑜

2

t time, t ⇒ 𝑣𝑣𝑥𝑥2 = 𝑣𝑣𝑜𝑜𝑜𝑜

2
+ 2𝑎𝑎𝑥𝑥 𝑥𝑥 − 𝑥𝑥𝑜𝑜 (3)
Total area under the curve =total
change in position, i.e. displacement

19
Example: using calculus in constant acceleration
The example below show how calculus is used to find
velocity of constant acceleration motion
Equation 2:
1
𝑥𝑥 = 𝑥𝑥𝑜𝑜 + 𝑣𝑣𝑜𝑜𝑜𝑜 𝑡𝑡 + 𝑎𝑎𝑥𝑥 𝑡𝑡 2
2
Instantaneous velocity
𝑑𝑑𝑑𝑑 𝑑𝑑 1
𝑣𝑣𝑥𝑥 = = (𝑥𝑥𝑜𝑜 + 𝑣𝑣𝑜𝑜𝑜𝑜 𝑡𝑡 + 𝑎𝑎𝑥𝑥 𝑡𝑡 2 )
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 2
𝑑𝑑 𝑑𝑑 𝑑𝑑 1
= 𝑥𝑥𝑜𝑜 + 𝑣𝑣𝑜𝑜𝑜𝑜 𝑡𝑡 + ( 𝑎𝑎𝑥𝑥 𝑡𝑡 2 )
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 2
𝑣𝑣𝑜𝑜𝑜𝑜 𝑑𝑑
0 (1/2)𝑎𝑎𝑥𝑥 𝑡𝑡 2
𝑑𝑑𝑑𝑑
1
= 𝑣𝑣𝑜𝑜𝑜𝑜 + 𝑎𝑎𝑥𝑥 2𝑡𝑡
2
𝑣𝑣𝑥𝑥 = 𝑣𝑣𝑜𝑜𝑜𝑜 + 𝑎𝑎𝑥𝑥 𝑡𝑡 (Eq. 1)

20
Example
𝑥𝑥 = 100 + 𝑡𝑡 2 + 3𝑡𝑡 𝑑𝑑𝑑𝑑
𝑣𝑣𝑥𝑥 = = 2𝑡𝑡 + 3
𝑑𝑑𝑑𝑑
10000 200

8000
150

6000
100
4000

50
2000

0 0
0 20 40 60 80 100 0 20 40 60 80 100
Time, t (sec) Time, t (sec)

𝑑𝑑𝑣𝑣𝑥𝑥
𝑎𝑎𝑥𝑥 = =2
𝑑𝑑𝑑𝑑

21
 Chapter 2
Part 2:
Motion in Two or Three Dimensions
Topics for Chapter 2 Part 2
 Use a vector r⃗ to represent the position of a body
 the velocity vector v
 the acceleration vector of a body a
 the curved path of a projectile
 circular motion

23
Introduction
 Someone throws a baseball. What determines where the
baseball lands?
 If a cyclist is going around a curve at constant speed, is
he accelerating?
 We need to extend our description of motion to two and
three dimensions.

24
Position Vector

 Position of a point P (an object)

is given by the position
vector r⃗
 The position vector from the
P
origin to point P has
components x, y, and z. 𝑟𝑟 𝑧𝑧𝑘𝑘�
 These components are the
coordinates of P 𝑥𝑥 𝚤𝚤̂

 Motion is the position vector as 𝑦𝑦𝚥𝚥̂

a function of time

r t = x t ı̂ + y t ȷ̂ + z(t)k�

25
Average velocity
z
The average velocity (a vector) Position at time 𝑡𝑡2
between two points is the
displacement (a vector) divided 𝑟𝑟2 ∆𝑟𝑟
by the time interval between
Position at time 𝑡𝑡1
the two points, and it has the 𝑟𝑟1
same direction as the y
displacement. x r1 = x1 ı̂ + y1 ȷ̂ + z1 k�
r2 = x2 ı̂ + y2 ȷ̂ + z2 k�
∆r ∆x ∆y ∆z ∆r = ∆xı̂ + ∆yȷ̂ + ∆zk�
vav = = ı̂ + ȷ̂ + k�
∆t ∆t ∆t ∆t
1
= x2 − x1 ı̂ + y2 − y1 ȷ̂ + z2 − z1 k�
(t2 −t1 )

26
Instantaneous Velocity
 The instantaneous velocity v is
the instantaneous rate of change Instantaneous velocity is
of position vector r with respect y tangential to the path
dr 𝑣𝑣𝑦𝑦 𝑣𝑣
to time. It is a vector: v =
dt
𝑣𝑣𝑥𝑥
 The components of the
instantaneous velocity are
dx dy dz 𝑟𝑟
vx = , vy = , vz =
dt dt dt
 The instantaneous velocity of a x
particle is always tangent to its
path. ∆r d⃗r
v ≡ lim vav = lim =
t→0 t→0 ∆t dt
d⃗r dx dy dz
v= = ı̂ + ȷ̂ + k� = vx ı̂ + vy ȷ̂ + vz k�
dt dt dt dt

27
Average acceleration
The average acceleration during a time interval Δt is defined
as the velocity change during Δt divided by Δt.

𝑣𝑣1 𝑣𝑣1

𝑣𝑣2
𝑣𝑣2 ∆𝑣𝑣

∆v ∆vx ∆vy ∆vz

aav = = ı̂ + ȷ̂ + k�
∆t ∆t ∆t ∆t

28
Instantaneous acceleration
 The instantaneous acceleration
𝑣𝑣 (𝑡𝑡)
is the instantaneous rate of change A 𝑣𝑣 (𝑡𝑡)
of the velocity with respect to time.
 Any particle following a curved path B
is accelerating, even if it has 𝑣𝑣 (𝑡𝑡 + ∆𝑡𝑡)
constant speed, because the ∆𝑡𝑡
direction is changing. x
 The components of the
instantaneous acceleration are v
dvx dvy dvz
ax = , ay = , az = .
dt dt dt

∆v dv
a = lim = P
t→0 ∆t dt
dv dvx dvy dvz
a = = ı̂ + ȷ̂ + k�
dt dt dt dt
t
a = ax ı̂ + ay ȷ̂ + az k�
29
Direction of the acceleration Vector
Any particle following a curved path:
 Both the magnitude (speed) and the direction can change
 The direction of the acceleration vector depends on
whether the speed is constant, increasing, or decreasing,

𝑣𝑣 𝑣𝑣 𝑣𝑣
𝑎𝑎𝑥𝑥
𝑎𝑎𝑥𝑥
𝑎𝑎 𝑎𝑎𝑦𝑦
𝑎𝑎 𝑎𝑎𝑦𝑦
𝑎𝑎

Speed is constant: Speed is increasing: Speed is decreasing:

acceleration is acceleration is acceleration is pointing
perpendicular to velocity pointing forward backward

30
Determining motion in 2 and 3 dimensions.

 For 2 D and 3 D cases, we need to use vectors to

describe motion
 It is convenient to express the vectors with their
components, in terms of the unit vectors.
─ i.e. resolve the vectors into components along the
unit vectors
 Each component can be treated as one-dimension
motion.
 The velocity and acceleration of each component can
be calculated using one dimension equations.

31
Typical 2-D motion-Projectile motion
 An object or particle (called a projectile) is thrown near the earth's
surface, and it moves along a curved path under the action of gravity
only
 A projectile motion is a typical 2 dimension motion
 The trick: separate the motion into the motion along x and y directions
(the x and y components of the position vector)

32
What is a projectile motion?
 A projectile is any body given an initial velocity that then
follows a path determined by the effects of gravity and air
resistance.
 Begin by neglecting air resistance, and the curvature and
rotation of the earth. We consider gravity effect only.
Gravity gives an acceleration g.

33
The x and y Motion are separable
 In a projectile motion, the vertical motion is not affected by
horizontal motion. The two motions can be separated
 The horizontal position of the ball is the same as a ball moving
with a constant velocity with zero acceleration
─ Initial conditions (t = 0):
x0 = 0, y0 = 0
vox = vo cos θo and voy = vo sin θ0
─ Horizontal motion: 𝑣𝑣𝑥𝑥 = 𝑣𝑣𝑜𝑜𝑥𝑥
x = 0 + vox t = vo cos θo t
─ Vertical motion:
vy = vo sin θo − gt
1 2
y = 0 + voy t − gt
2
1 2
y = (vo sin θo )t − gt
2
34
Equations for projectile motion
We apply the four equations to
1) the x direction motion (constant velocity, 𝑎𝑎𝑥𝑥 = 0)
2) the y direction motion (constant acceleration, 𝑎𝑎 = 𝑎𝑎𝑦𝑦 = −𝑔𝑔 )

vx = vox vy = voy + ay t
x = xo + vox t 1
y = yo + voy t + ay t 2
vx2 = vox
2 2

vox + vx vy2 = voy

2
+ 2ay y − yo
x − xo = t
2 voy + vy
y − yo = t
2
= vox t

35
Circular motion: another example of 2-D motion
 Another important example of two dimensional motion is
the uniform circular motion
 Examples: the orbit of satellite around the earth, a car goes
around a circular roundabout or corner.

36
Uniform circular motion—Figure 3.27
 For uniform circular motion, the speed is constant and
the acceleration is perpendicular to the velocity.
 It is called centripetal (center seeking) acceleration

 Velocity:
─ Magnitude: constant v
─ The direction of the velocity is tangent
to the circle 𝑟𝑟
 Acceleration:
v2
─ Magnitude: a = r
─ toward the center of the circle of motion
 Period:
─ time interval required for one complete
2πr
revolution of the particle 𝑇𝑇 =
v

37
Derivation of the centripetal acceleration
 For a uniform circular motion:
∆s v∆t
∆θ = =
r r
v(v∆t)
∆v = ∆v = v∆θ =
r
∆v v 2
= = average acceleration
∆t r
 Instantaneous acceleration
∆v v 2
𝑎𝑎 = lim =
𝑡𝑡→0 ∆t r
 Instantaneous acceleration =
average acceleration in uniform
circular motion as it is a constant

38
Use calculus to find centripetal acceleration

s sin 𝜃𝜃

cos 𝜃𝜃

r = x𝚤𝚤̂ + y𝚥𝚥̂
𝑠𝑠
𝜃𝜃 = = 𝜔𝜔𝜔𝜔; 𝜔𝜔 is the angular frequency
𝑟𝑟
𝑑𝑑𝑑𝑑 1 𝑑𝑑𝑑𝑑 v
𝜔𝜔 = = =
𝑑𝑑𝑑𝑑 𝑟𝑟 𝑑𝑑𝑑𝑑 r

The angle between the vector r and the x-axis is 𝜃𝜃 = 𝜔𝜔𝜔𝜔

r = x𝚤𝚤̂ + y𝚥𝚥̂ = r cos(𝜃𝜃)𝚤𝚤̂ + r sin(𝜃𝜃)𝚥𝚥̂ = r cos(𝜔𝜔𝑡𝑡)𝚤𝚤̂ + r sin(𝜔𝜔𝑡𝑡)𝚥𝚥̂
dr
v = = −r ωsin(𝜔𝜔𝜔𝜔)𝚤𝚤̂ + r𝜔𝜔 cos(𝜔𝜔𝜔𝜔)𝚥𝚥̂
dt
Direction of v ⊥ r
dv
a = = −r ω2 cos 𝜔𝜔𝜔𝜔 𝚤𝚤̂ − rω2 sin 𝜔𝜔𝜔𝜔 𝚥𝚥̂ = −ω2 𝑟𝑟
dt
39
Review: Equations
Equations of linear motion Projectile motion:
with constant acceleration ─ Horizontal motion: vx = vox
vx = vox + ax t (1) x = 0 + vox t = vo cos θo t
1 ─ Vertical motion:
x = xo + vox t + ax t 2 (2)
2 vy = vo sin θo − gt
vx2 = vox
2
+ 2ax x − xo (3) 1 2
y = (vo sin θo )t − gt
x − xo =
vox +vx
t (4) 2
2
Circular motion:
2- and 3-D motion: v2
Acceleration: a =
r t = x t ı̂ + y t ȷ̂ + z(t)k� r
2πr
d⃗r dx dy dz Period: 𝑇𝑇 =
v= = ı̂ + ȷ̂ + k� v
dt dt dt dt r = x𝚤𝚤̂ + y𝚥𝚥̂ = r cos(𝜔𝜔𝑡𝑡)𝚤𝚤̂ + r sin(𝜔𝜔𝑡𝑡)𝚥𝚥̂
= vx ı̂ + vy ȷ̂ + vz k� dr
v = = −r ωsin(𝜔𝜔𝜔𝜔)𝚤𝚤̂ + r𝜔𝜔 cos(𝜔𝜔𝜔𝜔)𝚥𝚥̂
dv dvx dvy dvz dt
a = = ı̂ + ȷ̂ + k� 𝐝𝐝v
dt dt dt dt a = = −r𝜔𝜔2 cos 𝜔𝜔𝜔𝜔 𝚤𝚤̂ − rω2 sin 𝜔𝜔𝜔𝜔 𝚥𝚥̂
= ax ı̂ + ay ȷ̂ + az k� 𝐝𝐝𝐝𝐝
= −𝜔𝜔2 r
40
Interesting Video quiz

If they are not shown in the lecture, you can watch the
videos on the web.
http://media.pearsoncmg.com/aw/aw_0media_physics/vt
d/video2.html
http://media.pearsoncmg.com/aw/aw_0media_physics/vt
d/video5.html
http://media.pearsoncmg.com/aw/aw_0media_physics/vt
d/video7.html#
http://media.pearsoncmg.com/aw/aw_0media_physics/vt
d/video9.html

41
Examples
Comments on the examples.
 The examples are chosen to show how the concepts learned
in this chapter are applied to problems.
 There is an example here about circular motion. You can
learn acceleration in circular motion in this example.
 One example about how to apply the equations for 1
dimensional motion with constant acceleration and velocity.
 Two examples about 2 dimension motion. You can learn
using vectors to describe 2 dimensional motion and the
concepts of average and instantaneous velocity and
acceleration.
 One example about projectile motion. You can learn how
the x and y motion in the projectile are treated separately
using 1 dimensional equations.

43
Centripetal acceleration on a curved road
A sports car has a lateral acceleration as its rounds a curve in the road.

44
 v2
a=
r

45
Two bodies with different accelerations

46
Calculating average and instantaneous velocity

A rover moves on the surface of Mars.

52
53
54
55
∆𝑟𝑟⃑

56
Calculating average and instantaneous acceleration

• Return to the Mars rover.

• Follow Example 3.2.

59
60
 2 2
Magnitude of the average acceleration= 𝑎𝑎𝑎𝑎𝑎𝑎−𝑥𝑥 + 𝑎𝑎𝑎𝑎𝑎𝑎−𝑦𝑦 = −0.5 2 + 0.15 2

=0.522 m/s2

61
62
63
Height and range of a projectile

 A baseball is batted at an angle.

 Follow Example 3.7.

64
(c) The range is when the ball hits
the ground or when y=0 again, so
1
0 = 0 + 𝑣𝑣𝑜𝑜𝑜𝑜 𝑡𝑡2 + −𝑔𝑔 𝑡𝑡22
2
𝑡𝑡2 = 0 𝑜𝑜𝑜𝑜 = 2𝑣𝑣𝑜𝑜𝑜𝑜 /𝑔𝑔
2 29.6
𝑡𝑡2 = = 6.04𝑠𝑠
9.8
𝑅𝑅 = 𝑣𝑣𝑜𝑜𝑜𝑜 𝑡𝑡2 = 22.2 6.04 𝑚𝑚 = 134𝑚𝑚