Motion in a Plane
Chapter 3
Physics Chapter 3
1
Position Vector
• Goes from the origin to the object – point p
y
P

r
Physics Chapter 3
x
2
Average Velocity Vector
• Change in displacement over change in
time

v av 

r
t
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Instantaneous velocity vector

v  lim
t  0

r
t
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Instantaneous velocity
• In two dimensions

v v
2
vx  vy
tan  
2
vy
vx
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Average acceleration vector

a av 

v
t
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Instantaneous acceleration
vector

a  lim
t  0

v
t
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Example
• A particle has x and y coordinates
(4.0 m, 2.0 m) at time t1 = 2.0 s and
coordinates (7.0 m, 6.0 m) at time t2 = 2.5
s. Find the components of the average
velocity and the magnitude and direction of
the average velocity during this time
interval.
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Projectile Motion
• Projectile – any body given an initial
velocity which then follows a path
(trajectory) based on gravitational
acceleration and air resistance
• Thrown ball
• Bullet
• Dropped package
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Projectile motion
• We will neglect air resistance
• We will also neglect curvature and rotation
of the earth
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Projectile Motion
• Consists of both horizontal and vertical
motion
• We will break these problems into x and y
components to make them easier to solve
Physics Chapter 3
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Projectile motion
• Two dimensional – gravity only acts
vertically
• We assign y as the vertical direction
• We assign x as the horizontal direction
ax  0
ay  g
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Projectile Motion
0
v x  v0 x  a x t
v x  v0 x
v y  v 0 y  gt
x  x0  v0 xt 
1
2
0
a xt
2
x  x0  v xt
y  y0  v0 yt 
Physics Chapter 3
1
gt
2
2
13
Projectile Motion
• If we express initial velocity in terms of its
magnitude and angle with the x-axis
v 0 x  v 0 cos  0
v 0 y  v 0 sin  0
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Projectile Motion
• We can calculate the projectile’s speed and
the direction of its velocity
v
2
vx  vy
tan  
2
vy
vx
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Trajectory shape
• Projectiles always travel in parabolas
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Example
• A policeman chases a thief across city
rooftops. They are both running at 5 m/s
when they come to a gap between
buildings that is 4 m wide and has a drop
of 3 m.
• The thief leaps at 5 m/s at an angle of 45°.
Does he clear the gap?
• The policeman leaps at 5 m/s horizontally.
Does he clear the gap?
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Example
x0
y0
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Example – the thief
x  x0  v xt
x0  0
x  v xt
v0  5
m
s
@ 45

 m

vx   5
cos
45

 s 
 m 

x   5
 cos 45  t
 Chapter 3 
  sPhysics
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Example – the thief
y0  0
v0  5
m
@ 45

v0 y
s
0
y  y0  v0 yt 
y  v0 y t 
1
1
 m

 5
sin
45

 s 
gt
2
2
gt
2
2
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Example – the thief
1
2
gt  v 0 y t  y  0
2
when he hits the roof,
y  3 m
m  2  m 


 4 .9 2  t    5
 sin 45  t  3 m  0
s 

 s 

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Quadratic formula
ax  bx  c  0
2
x
b
b  4 ac
2
2a
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Example – the thief
m  2  m 


 4 .9 2  t    5
 sin 45  t  3 m  0
s 

 s 

t

m

   5
sin
45


s 


2

m
m


 3 m 

5
sin
45

4
4
.
9


2 


s 
s 



m

2  4 .9 2 
s 

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Example – the thief
t
m

 3.54

s 

2
2

m  
m 
 12 . 5 2    58 . 8 2 
s  
s 

m

 9 .8 2 
s 

m 
m

 3.54
   8.44

s  
s 

t
m

 9 .8 2 
s 

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Example – the thief
11 . 98
t
m
 4 .9
m
s or t 
s
m
m


 9 .8 2 
 9 .8 2 
s 
s 


t  1 . 22 s
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Example – the thief
 m 

x   5
 cos 45  t
 s 

 m 

x   5
 cos 45  1 . 22 s 
 s 

x  4 . 32 m
• Yes, he clears the gap
Physics Chapter 3
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Example – the policeman
• Everything is the same, except that
v0  5
m
@0

s
 m
vx   5

 s 
v0 y  0
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Example – the policeman
 m
x  5
t
 s 
y
1
gt
2
2
y
t 
2

1
g
2
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Example – the policeman
when he hits the roof,
y  3 m
t 
2
3m
m

  4 .9 2 
s 

t  0 . 61 s
2
2
t  0 . 78 s
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Example – the policeman
 m
x  5
t
 s 
 m
x  5
 0 . 78 s 
 s 
x  3 .9 m
• No, he does not clear the gap
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Example 2
•
A rookie quarterback throws a football
with an initial upward velocity
component of 16.0 m/s and a horizontal
component of 20.0 m/s.
a) How much time is required for the
football to reach the highest point of the
trajectory?
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Example 2 a)
• At the highest point, vy must be zero.
• Before that, it was moving up, and after
that it is moving down, so it must
momentarily stop at that point.
v y  v 0 y  gt
vy 0
v 0 y  16 . 0
Physics Chapter 3
m
s
32
Example 2 a)
m

0  16 . 0
  9 .8 2 t
s 
s 
m
16 . 0
t 
m
s
m

 9 .8 2 
s 

t  1 .6 s
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Example 2 b)
• b)How high is this point?
y  y0  v0 yt 
y0  0
1
gt
2
2
v 0 y  16 . 0
m
s
t  1 .6 s
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Example 2 b)
m
1
m

2
y   16
 1 . 6 s    9 . 8 2  1 . 6 s 
s 
2
s 

y  25 . 6 m  12 . 5 m
y  13.1 m
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Example 2 c)
•
c) How much time is required for the ball
to return to its original height?
y  y0  v0 yt 
1
gt
2
2
y0
y0  0
v 0 y  16 . 0
m
s
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Example 2 c)
0  v0 y t 
1
2
1
gt
2
2
gt  v 0 y t  0
2
1

t  gt  v 0 y   0
2

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Example 2 c)
1

or  gt  v 0 y   0
2

t  0
1
2
gt  v 0 y
t 
2v0 y
g
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Example 2 c)
m

2  16

s 

t
m
9 .8 2
s
t  3 .2 s
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Example 2 d)
•
d) How does the answer to part c)
compare to the answer to part a)?
•
It is double. When something is thrown
upward, it always takes the same amount
of time to go up as it does to come down.
So the total flight time (back to the
original height) is always twice the time
to the highest point.
•
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Example 2 e)
•
e) How far has the football traveled
horizontally?
x  x0  v xt
x0  0
v x  20
m
s
t  3 .2 s
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Example 2 e)
m

x   20
 3 . 2 s 
s 

x  64 m
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Normal acceleration
• Normal means perpendicular
• The component of acceleration that is
perpendicular to the path
• Shows change in direction of velocity
a  or a norm
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Tangential acceleration
• Parallel to the path
• Shows change in magnitude of velocity
(change in speed)
a
or a tan
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Uniform Circular Motion
• When an object moves around a circle with
constant speed, the acceleration vector
points towards the center of the circle.
• The velocity is tangent to the circle. In
order for the speed to remain constant, the
acceleration must be towards the center of
the circle.
• We call this centripetal acceleration.
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Uniform circular motion
• See Fig. 3-11 on page 63
v

v1
a av 
s
v 
R
v
t
v1
s
R

v1  s
a
Rt
v
R
a 
v
lim
t0
s
t
2
R
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Uniform circular motion
• The period of any motion is the time it takes to
make one complete cycle. It is denoted with the
letter T.
• In this case it’s the time to go around the circle
once.
v av 
d
v
t
2 R
T
4 R
2
a 
T
2
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Example
• The radius of the earth’s orbit around the
sun is 1.49 x 1011 m, and the earth travels
around the sun in 365.25 days.
• What is the earth’s orbital speed in m/s?
• What is the acceleration of the earth
toward the sun in m/s2?
Physics Chapter 3
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