Transcript PowerPoint

Chapter 3
Kinematics in Two Dimensions
GT Homework 2012
•
The treasure map in the figure gives the following directions to the buried
treasure: “Start at the old oak tree, walk due north for 540 paces, then due east
for 110 paces. Dig.” But when you arrive, you find an angry dragon just north of
the tree. To avoid the dragon, you set off along the yellow brick road at an angle
60o EN. After walking 400 paces you see an opening through the woods.
In which direction and how far do you need to
go to get to the treasure from there?
Average velocity is the
displacement divided by
the elapsed time.
 

 r  ro r
v

t  to t


r
v  lim
t 0 t
The instantaneous velocity indicates how fast
the car moves and the direction of motion at each
instant of time.


r
v  lim
t 0 t



v PG  v PT  vTG
Example 11
Crossing a River
The engine of a boat drives it across a river that is 1800m wide.
The velocity of the boat relative to the water is 4.0m/s directed
perpendicular to the current. The velocity of the water relative
to the shore is 2.0m/s.
(a) What is the velocity of the
boat relative to the shore?
(b) How long does it take for
the boat to cross the river?



vBS  vBW  v WS
vBS  v
2
BW
v
2
WS

4.0 m s   2.0 m s 
 4.5 m s
 4.0 

  tan 

63

 2.0 
1
1800m
t
 450s
4.0m s
2
2
DEFINITION OF AVERAGE ACCELERATION
 

 v  v o v
a

t  to
t

v

v

vo
Equations of Kinematics – know by heart!
v  vo  at
x  12 vo  v  t
v  v  2ax
2
2
o
x  vot  at
1
2
2
Projectiles
• Projectiles move in 2 directions:
1) HORIZONTALLY – CONSTANT
VELOCITY
2) VERTICALLY – FREE-FALL
(CONSTANT ACCELERATION)
3) AIR RESISTANCE IS NEGLIGIBLE IN
EACH CASE
HORIZONTAL LAUNCH
Under the influence of gravity alone, an object near the
surface of the Earth will accelerate downwards at 9.80m/s2.
ay  9.80m s
2
ax  0
vx  vox  constant
For a horizontally launched projectile,
vyi = 0
A Falling Care Package
The airplane is moving horizontally with a constant velocity
of +115 m/s at an altitude of 1050m. Determine its horizontal
displacement from the place of the drop
Given
Dy = 1050m
vx = 115m / s
g = 9.8m / s 2
Dx
U
Equations / Substitutions
Horizontal
Dx = vx Dt
Dt = ?
Dx = vx
2Dy
g
Solution
Vertical
1
Dy = g(Dt)2
2
2Dy
Dt =
g
2100
Dx = 115
9.81
2100
Dx = 115
9.81
Dx = 1.7 ×10 3 m
3.3 Projectile Motion
Conceptual Example 5
I Shot a Bullet into the Air...
Suppose you are driving a convertible with the top down.
The car is moving to the right at constant velocity. You point
a rifle straight up into the air and fire it. In the absence of air
resistance, where would the bullet land – behind you, ahead
of you, or in the barrel of the rifle?
Lab
• “Hunger Games”.
3.3 Projectile Motion
Example 6 The Height of a Kickoff
A placekicker kicks a football at and angle of 40.0 degrees and
the initial speed of the ball is 22 m/s. Ignoring air resistance,
determine the maximum height that the ball attains.
vo

voy
vox
voy  vo sin   22m ssin 40  14m s

vox = vo cosq = ( 22m s) cos 40 = 17m s
Given
vi = 22m / s
q = 40.0
g = 9.8m / s 2
Dx = ?
Dy = ?
U
Equations / Substitutions
Solution
Horizontal
Vertical
vi = 17m / s
vi = 14m / s
-viy
Dt =
g
Dx = vx Dt
Dt = ?
Dy =
( )
- viy
2g
2
-14
Dt =
= 1.43s
-9.8
Dx = 2.9 ×17
Dx = 49(m)
- (14 )
2(-9.8)
Dy = 10(m)
2
Dy =
Conceptual Example 10
Two Ways to Throw a Stone
From the top of a cliff, a person throws two stones. The stones
have identical initial speeds, but stone 1 is thrown downward
at some angle above the horizontal and stone 2 is thrown at
the same angle below the horizontal. Neglecting air resistance,
which stone, if either, strikes the water with greater velocity?
Rocket Lab