Transcript PowerPoint Presentation - Chapter 3 Kinematics in 2d
Chapter 3
Motion in 2 dimensions
1) Displacement, velocity and acceleration
• displacement is the
vector
from initial to final position
r
r
r
r
r
r 0
r x
r y
x
- component of
r
r
y
- component of
r
r
x
y
y
x
r x
y
r y
r
0
x
r
0
y
x
x
0
y
y
0
x
1) Displacement, velocity and acceleration v
• average velocity
v
r
r
t v x
v y
x
t
y
t
x y
x
0
t
y
0
t
1) Displacement, velocity and acceleration v v
0 • instantaneous velocity
v
lim
t
0
t r
r
v x v y
lim
t
0
x
lim
t
0
t
y
t
v
is tangent to the path
v
can change even if v is constant
1) Displacement, velocity and acceleration v v a v
0 • average acceleration
a
r
v
t
r
v
t v
r 0
a x
a y
v x
t
v y
t not in general parallel to velocity
1) Displacement, velocity and acceleration v v
0 • instantaneous acceleration
a
lim
t
0 r
v
t a x a y
lim
t
0 lim
t
0
v x
t
v y
t
1) Displacement, velocity and acceleration v a v
r
v v
0 • instantaneous acceleration
a
lim
t
0 r
v
t
- object speeding up in a straight line
v
0
v
r
v acceleration parallel to velocity
- object at constant speed but changing direction
acceleration perp. to velocity
2) Equations of kinematics in 2d
• Superposition (Galileo): If an object is subjected to two separate influences, each producing a characteristic type of motion, it responds to each without modifying its response to the other.
• That is, we consider
x
and
y
motion separately
2) Equations of kinematics in 2d v y v x v x v y
A bullet fired vertically in a car moving with constant velocity, in the absence of air resistance (and ignoring Coriolis forces and the curvature of the earth), will fall back into the barrel of the gun. That is, the bullet’s
x
-velocity is not affected by the acceleration in the
y
-direction.
2) Equations of kinematics in 2d
• That is, we can consider
x
and
y
motion separately
2) Equations of kinematics in 2d
• That is, we can consider
x
and
y
motion separately
Displacement
:
x
,
y
(
x
0 0,
y
0 0)
Velocity
:
v x
,
v y v
0
x
,
v
0
y Acceleration
:
a x
,
a y
(
a
0
x
a x
,
a
0
y Time
:
t
(
t
0 0)
a y
)
x v x
v
0
x t v
0
x v x
2
x
v
0
x
1 2 (
v
0
x
2
x
v x t
a x t
1 2 1 2
a
2
x a v x
)
t a x t
2
t
2
x x y v y
v
0
y t v
0
y
1 2
a y t
2
a y t v y
2
y
v
0
y
2 2
a y y
1 2 (
v
0
y
v y
)
t y
v y t
1 2
a y t
2
Example y v
0
a
Find
x
,
x
v
0
x y
, r
v
at
t t
1 2
a x
t
2 7.0 s 740 m
y
v
0
y t
1 2
a y t
2 390 m
v x v y v
v
0
x
v
0
y
a x t
a y t
190 m/s 98 m/s
v x
2
v y
2 210 m/s
v x
tan
v v x y
0.516
27º
v v
0
x
22 m/s;
v
0
y
14 m/s
a x
24 m/s 2 ;
a y t
7.0 s 12 m/s 2
v
0
v
0
x
2
v
0
y
2 26 m/s tan
v y v x
0.516
32.5º
3) Projectile Motion (no friction)
a) Equations Consider horizontal (
x
) and vertical (
y
) motion separately (but with the same time) Horizontal motion: No acceleration ==>
a x
=0
x v x
v
0
x t v
0
x
1 2
a x a x t
2
t
v x x
v
0
x t
v
0
x
Vertical motion: Acceleration due to gravity ==>
a y
= ±
g - usual equations for constant acceleration
3) Projectile Motion (no friction)
Example: Falling care package Find
x.
Step 1
: Find
t
from vertical motion
Step 2
: Find
x
from horizontal motion
x
3) Projectile Motion (no friction)
Example: Falling care package
Step 1:
Given
a y , y, v 0y t
Solve
y
v
0
y t
1 2
a y t
2 1 2
gt
2 2
y
14.6 s
g x
3) Projectile Motion (no friction)
Example: Falling care package
Step 2: x
v
0
x t
1680 m
x
3) Projectile Motion (no friction)
b) Nature of the motion: What is
y(x)
?
Eliminate
t
from
y(t)
and
x(t): y x x
v
0
x t
;
y
v
0
y t
1 2
gt
2
t
x v
0
x
Substitute:
y
v
0
y
v
0
x
x
g
2
v
0
x
2
x
2
y(x)
is a parabola
3) Projectile Motion (no friction)
c) Cannonball physics Initial velocity r
v
is given: Either (
v
0
x
,
v
0
y
) or (
v
0 , ) Find (i) height, (ii) time-of-flight, (iii) range
3) Projectile Motion (no friction)
Initial velocity r
v
is given: c) Cannonball physics Either (
v
0
x
,
v
0
y
) or (
v
0 , ) (i) Height: Consider only
y
-motion:
v 0y
given,
a y =-g
known Third quantity from condition for max height:
v y =0
Use
v
2
v
0
y
2 2
a y y
When
y
H
,
v
0, so 0
v
0
y
2 2
gH
H
v
0
y
2 2
g
3) Projectile Motion (no friction)
Initial velocity r
v
is given: c) Cannonball physics Either (
v
0
x
,
v
0
y
) or (
v
0 , ) (ii) Time-of-flight: Consider only Use
y
v
0
y t
y
1 2 -motion:
a y t
2
v 0y
given,
a y =-g
known Third quantity from condition for end of flight;
y=0
When
y
0, 0
v
0
y t
1 2
gt
2 (Trivial solution:
t
0) Non-trivial solution:
t
2
v
0
y g
3) Projectile Motion (no friction)
Initial velocity r
v
is given: c) Cannonball physics Either (
v
0
x
,
v
0
y
) or (
v
0 , ) (iii) Range: Consider
x
-motion using time-of-flight:
x=v 0x t
For
y
0,
x
R
and
t
2
v
0
y g
, so from
x
v
0
x t
,
R
2
v
0
x v
0
y g
or,
R
v
0 2 (2 cos sin )
g v
0 2 sin(2 )
g