Summary (CH1 and CH2)
•  Vectors decomposition
•  Vector addition

V = Vx iˆ + Vy ˆj =|V | cosθ iˆ + |V | sin θˆj

Vtot = (V1x + V2x )iˆ + (V1y + V2y ) ˆj
V1y + V2y
tgθ =
V1x + V2x

| Vtot |= (V1x + V2x ) 2 + (V1y + V2y ) 2
Average velocity/acceleration
•  1d motion (constant acceleration) v = Δx/Δt = (x-x )/(t-t )
avg
0
0
v x = v 0x + ax t
aavg= Δv/Δt = (vx-v0x)/(t-t0)
€
1 2
x = x 0 + v 0x t + ax t
2
2
v x2 = v 0x
+ 2ax (x − x 0 )
1
x − x 0 = (v x + v x 0 )t
2
Instantaneous velocity/acceleration
dx
dt
dv x
=
dt
vx =
ax
€
Goals for Chapter 3: 2D or 3D motions
•  position, velocity, and acceleration vectors (2D/3D)
•  projectile motion
–  Intial velocity + Gravitational acceleration
•  circular motion
–  Uniform circular motion
–  Non-uniform circular motion
•  relative velocity
Next class
Opening question
•  When the race car goes
around a curve, is it
accelerating? If it is, in
what directions?
•  If I drop a pen, does it hit
the ground at the same
time as one throw
horizontally?
Position relative to the origin
•  An overall position relative to the origin can have
components in x, y, and z dimensions.
•  Straight line motion is the exception, not the norm
•  The path for a particle is generally a curve.
Displacement and Average velocity (2d) vectors
•  The average velocity between two points will have the
same direction as the displacement.
Note: In straight line
motion,
Velocity is either
positive, or negative
In 2d-3d motion,
displacement and
velocity is defined by its
components in all 2(3)
directions

v = v xiˆ + v y ˆj

Δr = (Δx)iˆ + (Δy) ˆj


r2 + r1
Compare and graphically! Which is head to tail?


Which is tail to tail?
r2 − r1
Components of velocity(2D)
•  A velocity in the xy-plane can be decomposed into separate
x and y components.
1d
2d
Δx
Δt
dx
vx =
dt
v avg =
 

r
−
r
Δ
r

v avg = 2 1 =
t 2 − t1 Δt
 
Δr dr

v = lim =
dt
Δt → 0 Δt
 dx ˆ dy ˆ
v= i+ j
dt dt
€
€
€
€
dx
dt
dy
vy =
dt
vx =
Short Quiz
If the position components are know as a function of time
x(t), y(t) (and z(t) in 3d) are given
Then what do we know at any given time:
A. Position vector
B. Average velocity
C. Instantaneous velocity
D. Average acceleration
E. Instantaneous acceleration
F. All of the above
Instantaneous velocity—Example
•  The figure blow shows velocity and acceleration as time passes.
•  A rover’s x,y-coordinates vary with time:
x = 2.0m − (0.25m /s2 )t 2
y = (1.0m /s)t + (0.25m /s3 )t 3
•  What is the rover’s instantaneous velocity vector?
•  What the magnitude and direction €of this vector at t=2.0s?
Instantaneous velocity—Example
What is the rover’s instantaneous velocity vector?
What the magnitude and direction of this vector at t=2.0s?
2
x = 2.0m − (0.25m /s )t
2
y = (1.0m /s)t + (0.025m /s3 )t 3
€
 ˆ ˆ
v = v xi + v y j
dx
v x = = −(0.5m /s2 )t
dt
dy
v y = =1.0m /s + (0.075m /s3 )t 2
dt

⇒ v (t) = [−(0.5m /s2 )t]iˆ + (1.0m /s + (0.075m /s3 )t 2 ) ˆj

v (2.0s) = [−(0.5m /s2 )2.0s]iˆ + (1.0m /s + (0.075m /s3 )(2.0s) 2 ) ˆj

v (2.0s) = (−1.0m /s)iˆ + (1.3m /s) ˆj
1.3
tgα =
= −1.3
−1.0
α =180 o + tg−1(−1.3) =180 o + (−52 o ) =128 o

The acceleration vector a
•  The acceleration vector can result in a change in either the
magnitude OR the direction of the velocity. (Can both be
changed at the same time?What is your daily example?)
•  What is the direction of the acceleration of this race car?
•  Acceleration always has the same direction as the velocity
change
€
Or is this ΔV?
Instantaneous acceleration—Example
•  From the previous example, we can also work out the
instantaneous accelerations.
2
x = 2.0m − (0.25m /s )t
dx
v x = = −(0.5m /s2 )t
dt
dy
v y = =1.0m /s + (0.075m /s3 )t 2
dt
2
y = (1.0m /s)t + (0.025m /s3 )t 3
€
dv x
ax =
= −0.5m /s2
dt
dv y
ay =
= 0.15m /s3 t
dt

a(2.0s) = (−0.5m /s2 )iˆ + (0.3m /s2 ) ˆj
tgβ = −0.6
β =149 o
Parallel and perpendicular components of acceleration a// and a⊥
What does a// mean?
Changing the instantaneous speed
(magnitude of velocity)
What does a⊥ mean?
Changing the direction– turning.
€
€
Skiing through a valley (direction of accelerations)
•  What is the
direction of the
accelerations of the
skier at the various
points?
Normal is a vector perpendicular
to the tangent of a curve
Q3.2
This illustration shows the path
of a robotic vehicle, or rover.
What is the direction of the
rover’s average acceleration
vector for the time interval from
t = 0.0 s to t = 2.0 s?
A. up and to the left
B. up and to the right
C. down and to the left
D. down and to the right
E. none of the above
Q3.3
The motion diagram shows an object moving along a
curved path at constant speed. At which of the points A, C,
and E does the object have zero acceleration?
A. point A only
B. point C only
C. point E only D. points A and C only
E. points A, C, and E
Summary of Chapter 3, part 1
•  Transition from 1d straight line motion to 2d (3d) motions
1d
2d
Δx
Δt
dx
Vx =
dt
Vavg =
•  Decompositions of vectors (position, velocity, acceleration)

 dr dx ˆ dy ˆ dz ˆ

2
2
2
v=
= i+
j+ k
€| r |= r = x + y + z
dt dt
dt
dt


2
2
2
| v |= v = v x + v y + v z
 dv dv x ˆ dv y ˆ dv z ˆ
a=
=
i+
j+
k
dt
dt
dt
dt
vy
2
2
2
tan α =
d
x
d
y
d
zˆ

(2d)
ˆ
ˆ
a= 2 i + 2 j+ 2 k
vx
dt
dt
dt
€
€
Q3.7
The velocity and acceleration of an object at a certain instant are
At this instant, the object is
A. speeding up and following a curved path.
B. speeding up and moving in a straight line.
C. slowing down and following a curved path.
D. slowing down and moving in a straight line.
E. none of these is correct
Q3.8*
The velocity and acceleration of an object at a certain instant are
At this instant, the object is
A. speeding up and following a curved path.
B. speeding up and moving in a straight line.
C. slowing down and following a curved path.
D. slowing down and moving in a straight line.
E. none of these is correct
Projectile motion (very important)
•  A projectile is any body given an initial velocity that then
follows a path determined by the effects of gravity and air
resistance.
•  Begin neglecting resistance
•  The trajectory is parabolic
It is easier to separate X and Ymotion
•  The red ball is
dropped, and the
yellow ball is fired
horizontally as it is
dropped.
•  The strobe marks
equal time intervals.
•  Is X-intervals
constant?
•  Is Y-intervals
constant
The equations of projectile motion (constant g)
•  Projectile motion sets
xo = 0 and yo = 0 then
obtains the specific results
shown at right.
x = (v 0 cos α 0 )t
1 2
y = (v 0 sin α 0 )t − gt
2
v x = v 0 cos α 0
v y = v 0 sin α 0 − gt
What if xo and yo are not 0
€
The equations of projectile motion (constant g)
x = (v 0 cos α 0 )t
1 2
y = (v 0 sin α 0 )t − gt
2
v x = v 0 cos α 0
v y = v 0 sin α 0 − gt
If xo and yo are not 0
€
x = x 0 + (v 0 cos α 0 )t
1
y = y 0 + (v 0 sin α 0 )t − gt 2
2
v x = v 0 cos α 0
v y = v 0 sin α 0 − gt
€
Q3.10
A projectile is launched at a 30° angle above the horizontal.
Ignore air resistance. The projectile’s acceleration is greatest
A. at a point between the launch point and the high point of
the trajectory.
B. at the high point of the trajectory.
C. at a point between the high point of the trajectory and
where it hits the ground.
D. misleading question — the acceleration is the same (but
nonzero) at all points along the trajectory
E. misleading question — the acceleration is zero at all points
along the trajectory
Physics of darts throwing (Not in the book)
•  Typical dart speed:
20.0m/s
Dart’s trajectory: Parabolic projectile
•  Distance between
player and board:
x = x 0 + v x 0t
y = y 0 ⇒ v y 0 t − 0.5gt 2 = 0 ⇒ v y 0 = 0.5at
x − x 0 = 4.0m ⇒ v x 0 t = 4.0m
v y 0 = v 0 sin α,v x 0 = v 0 cos α
4.0m
•  Where should the
player aim?
v0=20m/s
α
y = y 0 + v y 0 t − 0.5gt 2
€
t=
4.0m
v 0 cos α
Yoffset
4.0m
v 0 cos α
4.0m
⇒ sin2α = g 2
v0
v 0 sin α = 0.5g
α = 2.8 o
4.0m
Should aim at Yoffset=4.0m * tg α = 19.6 cm
€
From darts to rifle sniper (will be in the exams)
•  Typical rifle bullet
speed: v0=1000.0m/s
•  Distance sniper and
target: (x-x0)=50.0m
•  Where should the
sniper aim?
A: 1.25cm higher
B: 2.5cm higher
C: 1.25cm lower
D: 2.5cm lower
x − x0
v 02
= tgα (x − x 0 )
sin2α = g
Yoffset
At very small angles sin2α ≈ 2sin α ≈ 2tgα
€
Yoffset
g(x − x 0 ) 2
=
2v 0 2
€
The slower (and/or farther) your
€
dart/bullet
travel,
the more you need to compensate for
gravity by aiming higher
(It takes longer to hit the target, therefore
it would have dropped more vertically)
The effects of wind resistance
•  Cumulative effects
can be large.
•  Peak heights and
distance fall.
•  Trajectories cease
to be parabolic.
Motion in a circle
•  The vehicle could be speeding up in the curve,
slowing down in the curve, or undergoing uniform
circular motion.
Finding motion information
 
a⊥v
•  What is the direction of
the instantaneous
acceleration if the
motion is along curve
with constant speed?


| Δv | | Δv |
Δs
aav =
,
=
Δt
v1
R
v Δs
⇒ aav = 1
R Δt
Because
€
€
€
v1 Δs
v
Δs
= 1 lim
Δt → 0 R Δt
R Δt → 0 Δt
Δs
lim
= v1
Δt → 0 Δt
v1 = v
arad = lim
Using units to derive the equation for acceleration
•  How many physics observables are needed to describe a uniform
circular motion?
• 
Radius (m)
• 
Speed (m/s)
• 
Period (s)
• 
All of the above
• 
Only two of the first three
How about expressing arad
in terms of v and T?
•  Unit of arad: m/s2
arad
v2/R
R/T2
•  To get m/s2, we can use (m/s)2/m, or m/s2
Therefore
arad
v2
R
∝
∝
R
T2
Q3.11
You drive a race car around a circular track of radius 100 m at a
constant speed of 100 km/h. If you then drive the same car
around a different circular track of radius 200 m at a constant
speed of 200 km/h, your acceleration will be
A. 8 times greater.
B. 4 times greater.
C. twice as great.
D. the same.
E. half as great.
Determination of the centripetal acceleration
•  Uniform circular motion and
projectile motion compared
and contrasted.
•  For uniform circular motion,
the acceleration is centripetal
•  Example: In a circular motion
with constant speed, what is
the magnitude of the
acceleration if R=5.0m and
2
4
π
R
T=4.0s?
a =
rad
T2
What is the radius if a=9.4m/s2,
and v=40m/s?
arad
v2
v2
€
= ⇒R=
R
arad
Q3.4
An object moves at a constant speed in a clockwise direction
around an oval track. The geometrical center of the track is at
point O. When the object is at point P, which arrow shows the
direction of the object’s acceleration vector?
A. #1 (directly away from O)
#2
#1
#3
P
#4
#5
O
B. #2 (perpendicular to the track)
C. #3 (in the direction of motion)
D. #4 (directly toward O)
E. #5 (perpendicular to the track)
Oval track
Q3.5
A pendulum swings back and forth, reaching a maximum angle of
45° from the vertical. Which arrow shows the direction of the
pendulum bob’s acceleration as it moves from left to right through
point Q (the low point of the motion)?
A. #1 (to the left)
B. #2 (straight up) 45°
45°
P
R
C. #3 (to the right)
#2
D. #4 (straight down)
E. misleading question — the
acceleration is zero at Q #1
Q
#3
#4
Q3.6
A pendulum swings back and forth, reaching a maximum angle of
45° from the vertical. Which arrow shows the direction of the
pendulum bob’s acceleration at P (the far left point of the motion)?
A. #1 (up and to the left)
B. #2 (up and to the right)
C. #3 (down and to the
right)
D. #4 (straight down)
E. #5 (down and to the
left )
#1
P
#2
45°
45°
R
#3
#5
#4
Q
Relative velocity – straight line and 3d
• Observer: Got off bus, and not moving
Bus: +5m/s
What is the Bus’s relative velocity to the observer?
• What if observer is running at +6m/s?
• Ever got scared in a car wash station?




• In general:
v A / B = v A − v B = −v B / A
 
v A , v B can be measured in any reference frame
Example:
€

v A = 2.0(m / s)iˆ + 4.5(m / s) ˆj

v B = 1.9(m / s)iˆ + 3.2(m / s) ˆj


vA / B = ?vB / A = ?
It is usually easier to express the velocity vectors in terms of
their x/y/z components, when dealing with relative velocity
€
Q3.12
The pilot of a light airplane with an airspeed of 200 km/h wants
to fly due west. There is a strong wind of 120 km/h blowing
from the north.
If the pilot points the nose of the airplane north of west so that
her ground track is due west, what will be her groundspeed?
A. 80 km/h
B. 120 km/h
C. 160 km/h
D. 180 km/h
E. It would impossible to fly due west in this situation.
Summary (CH3)

r = xiˆ + yˆj + zkˆ

 dr dx ˆ dy ˆ dz ˆ
v=
= i+
j+ k
dt dt
dt
dt

d
v
dv x ˆ dv y ˆ dv z ˆ

a=
=
i+
j+
k
dt
dt
dt
dt
 d 2 x ˆ d 2 y ˆ d 2z ˆ
a= 2 i + 2 j+ 2 k
dt
dt
dt
•  Position, velocity and
acceleration vectors in 3D
• 
Decomposition!
• 
Decomposition!
• 
Decomposition!
Circular motions:
v2
4π 2R
arad =
=
R
T2
Relative velocity:
•  Projectile motion
x = x 0 + (v 0 cos α 0 )t
1
y = y 0 + (v 0 sin α 0 )t − gt 2
2
v x = v 0 cos α 0
v y = v 0 sin α 0 − gt
€




v A / B = v A − v B = −v B / A
€
€