Lecture 5
 Goals: (Chapter 4.1-3)
 Introduce position, displacement, velocity and
acceleration in 2D
 Address 2D motion in systems with constant acceleration
(i.e. both magnitude and acceleration)
 Discuss horizontal range (special)
Key point 1: In many case motion occurs 2D with constant
acceleration (typically on the surface of a planet)
Key point 2: The “superposition principle” allows us to
discuss the x & y motion individually
Physics 201: Lecture 5, Pg 1
Decomposing vectors

Any vector can be resolved into components along the x and y axes
y
ry
(x,y)
r
vy

rx
(vx,vy)
vy
v
(ax,ay)
ay
ay
a
vx

vx
vx  v cos 
ax

ax
a x  a cos 
ry  y  r sin 

r  x î  y ĵ
v y  v sin 

v  vx î  v y ĵ
a y  a sin 

a  a x î  a y ĵ
r x y
v  v v
a  ax2  a y2
x
rx  x  r cos 
2
2
tan-1 ( y / x )
2
x
2
y
tan-1 ( vy / vx )
tan-1 ( ay / ax )
Physics 201: Lecture 5, Pg 2
Dynamics: Motion along a line but with a twist
(2D dimensional motion, magnitude and directions)

Particle motions involve a path or trajectory
In 2D the position of a particle is r = x i + y j
and this vector is dependent on the origin. (i , j unit vectors )

O’
ri ’
rf ’
Displacement, Dr, is
independent of origin
Physics 201: Lecture 5, Pg 3
Motion in 2D
 Position
 Displacement
 Velocity
(avg)


ri , ti and rf , t f
  
Dr  rf  ri

Dr

vavg 
Dt
Physics 201: Lecture 5, Pg 4
Instantaneous Velocity


As Dt  0 Dr shrinks and becomes tangent to the path
The direction of the instantaneous velocity is along a
line that is tangent to the path of the particle’s direction of
motion.


Dr dr

v  lim

Dt 0 Dt
dt
v
Physics 201: Lecture 5, Pg 5
Average Acceleration

The average acceleration of particle motion reflects
changes in the instantaneous velocity vector (divided by
the time interval during which that change occurs).

a avg

 v  v
Dv
f
i


Dt
Dt
Instantaneous
acceleration

Dv

a  lim
Dt 0 Dt

aavg
Physics 201: Lecture 5, Pg 6
Instantaneous Acceleration

Dv

a  lim
Dt 0 Dt

The instantaneous acceleration is a vector with components
parallel (tangential) and/or perpendicular (radial) to the
tangent of the path

Changes in a particle’s path may produce an acceleration
 The magnitude of the velocity vector may change
 The direction of the velocity vector may change
 Both may change simultaneously (depends: path vs time)
Physics 201: Lecture 5, Pg 7
2 D Kinematics

Position, velocity, and acceleration of a particle:
r= xi +y j
v = vx i + v y j
(i , j unit vectors )
a = ax i + ay j

d ( xiˆ  yĵ)
 dr
v 

dt
dt
dx
dy

î 
ĵ
dt
dt
 v x î  v y ĵ
x  x(Dt )
y  y (Dt )
dx
vx 
dt
d 2x
ax  2
dt
dy
vy 
dt
d2y
ay  2
dt
Physics 201: Lecture 5, Pg 8
2 D Kinematics (special case)
 If ax and ay are constant then
x ( Dt )  xi  v xi Dt  12 a x Dt 2
y ( Dt )  yi  v yi Dt  a y Dt
1
2
and if
2
ti  0
x (t )  xi  vxit  a x t
1
2
2
vx (t )  vxi  a x t
y (t )  yi  v yi t  12 a y t 2
v y (t )  v yi  a y t

In many case one of the a’s is zero
Physics 201: Lecture 5, Pg 9
Trajectory with constant
acceleration along the vertical
Special Case:
ax=0 &
ay= -g
vx(t=0) = v0
vy(t=0) = 0
vy(t) = – g t
Position
t = 0 x vs y
x(t) = x0 v0 t
y(t) = y0 - ½ g t2
y
4
What do the velocity and acceleration
vectors look like?
x
Physics 201: Lecture 5, Pg 10
Trajectory with constant
acceleration along the vertical
What do the velocity and acceleration
vectors look like?
Velocity vector is always tangent to the
curve!
t=0
y
Velocity
x vs y
4
x
Physics 201: Lecture 5, Pg 11
Trajectory with constant
acceleration along the vertical
What do the velocity and acceleration
vectors look like?
Position
Velocity
Velocity vector is always tangent to the
curve!
Acceleration may or may not be!
t=0
y
Acceleration
x vs y
4
x
Physics 201: Lecture 5, Pg 12
Another trajectory
Can you identify the dynamics in this picture?
How many distinct regimes are there?
0<t<3
 I.
3<t<7
7 < t < 10
vx = constant = v0 ; vy = 0
 II. vx = vy = v0
t=0
 III. vx = 0 ; vy = constant < v0
x vs y
What can you say about the
acceleration?
y
t =10
x
Physics 201: Lecture 5, Pg 13
Exercises 1 & 2
Trajectories with acceleration



A rocket is drifting sideways (from left to right) in deep
space, with its engine off, from A to B. It is not near any
stars or planets or other outside forces.
Its “constant thrust” engine (i.e., acceleration is constant) is
fired at point B and left on for 2 seconds in which time the
rocket travels from point B to some point C
 Sketch the shape of the path
from B to C.
At point C the engine is turned off.
 Sketch the shape of the path
after point C (Note: a = 0)
Physics 201: Lecture 5, Pg 14
Exercise 1
Trajectories with acceleration
B
From B to C ?
A.
B.
C.
D.
E.
A
B
C
D
None of these
A
B
C
B
B
C
C
B
C
D
C
Physics 201: Lecture 5, Pg 15
Exercise 2
Trajectories with acceleration
After C ?
A.
B.
C.
D.
E.
A
B
C
D
None of these
C
C
A
B
C
C
C
D
Physics 201: Lecture 5, Pg 16
Kinematics in 2 D; Projectile Motion

The position, velocity, and acceleration of a particle
moving in 2-dimensions can be expressed as:
r= xi +y j

v = vx i + vy j
Special Case: ax=0 & ay= -g
vx(t) = v0 cos 
vy(t) = v0 sin – g t
x(t) = x0 v0 cos t

a = ax i + ay j

v0

vy0
vx 0  v0 cos 
y(t) = y0 v0 sin t - ½ g t2
Physics 201: Lecture 5, Pg 17
Kinematics in 2 D; Horizontal Range

Given v0 and how far will and object travel horizontally?
Let
y0 = 0 = yinitial = yfinal x0 =0
Again: ax=0 & ay= -g
1. vx(t) = v0 cos 
2. vy(t) = v0 sin – g t
3. x(t) = 0 v0 cos t = R (range)

v0

vy0
vx 0  v0 cos 
4. y(t) = 0 = 0 v0 sin t - ½ g t2
4 gives: 0 = t (v0 sin - ½ g t)  t = 0, 2 v0 sin /g
R = v0 cos 2 v0 sin /g = v02 sin 2 / g
Maximum when dR/d = 0  cos 2 = 0 or 45°
Physics 201: Lecture 5, Pg 18
Parabolic trajectories ( v=10 m/s , g = - 10 m/s2)
90° R:0.0m H:5.0m t=2.00s
75° R:5.0m H:4.7m t=1.93s
60° R:8.7m H:3.7m t=1.73s
45° R:10.0m H:2.5m t=1.41s
30° R:8.7m H:1.2m t=1.00s
15° R:5.0m H:0.3m t=0.52s
Physics 201: Lecture 5, Pg 19
Example Problem
A medieval soldier is at the top of a castle wall. There are two
cannon balls. The 1st one he fires from the cannon and it
lands 200 m away. Simultaneously the 2nd cannon ball is
dislodged and falls directly down. The 2nd cannon ball lands
after 2.0 seconds and the fired cannon ball lands 2.0
seconds later. The ground is completely level around the
castle. At what angle from horizontal did he fire the cannon
(if g= -10 m/s2 & no air resistance)?
Physics 201: Lecture 5, Pg 20
Example Problem
A medieval soldier is at the top of a castle wall. There are two
cannon balls. The 1st one he fires from the cannon and it lands
200 m away. Simultaneously the 2nd cannon ball is dislodged
and falls directly down. The 2nd cannon ball lands after 2.0
seconds and the fired cannon ball lands 2.0 seconds later.
The ground is completely level around the castle. At what
angle from horizontal did he fire the cannon (if g= -10 m/s2 &
no air resistance)?
Find height of wall from 2nd cannon ball
0 = h + 0 – ½ 10 t2  h = 5(4) m = 20 m
Find angle
0 = h + v sin  (t+2) – ½ 10 (t+2)2  0 = 20 m + 4v sin  - 80 m
R = 200 m = v cos  (t+2)  4v = 200 / cos 
Combining 60 m = 200 m tan   16.7 degrees
Physics 201: Lecture 5, Pg 21
Example Problem
Example Problem: If the knife is thrown horizontally at 10 m/s
second and the knife starts out at 1.25 m from the ground,
then how far does the knife travel be for it hit the level
ground (if g= -10 m/s2 & no air resistance)?
Physics 201: Lecture 5, Pg 22
Example Problem
If the knife is thrown horizontally at 10 m/s second and the knife
starts out at 1.25 m from the ground, then how far does the
knife travel before it hits the level ground
(assume g= -10 m/s2 & no air resistance)?
 at t=0 the knife is at (0.0 m, 1.0 m) with vy=0
 after Dt the kinfe is at (x m, 0.0 m)
 x = x0 + vx Dt and y = y0 – ½ g Dt2
So x = 10 m/s Dt
and
0.0 m = 1.25 m – ½ g Dt2  2.5/10 s2 = Dt2
Dt = 0.50 s
x = 10 m/s 0.50 s = 5 .0 m
Physics 201: Lecture 5, Pg 23
Fini

For Thursday, Read all of Chapter 4
Physics 201: Lecture 5, Pg 24