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Chapter 3 Kinematics in Two or Three Dimensions
Vectors and Scalars
Vector Kinematics
Two Dimensional Motion with Constant Acceleration
Relative Velocity
Solving Problems Involving Projectile Motion
(Most common type of motions on earth)Projectile motion is motion with constant acceleration in two dimensions, where the acceleration is g and is downward (toward the earth).
3-7 Projectile MotionIf an object is launched at an initial angle of θ0 with the horizontal, the analysis is similar except that the initial velocity has a vertical component.
As shown in Figure, a ski-jumper leaves the ski track which is curved so that the jumper is projected upward at an angle from the end of the track with a speed of v0 m/s,. The landing incline below him falls off with a slope of tan . Where does he land on the incline?
D
D
tvx )cos( 0
20 2
1)sin( gttvy
xy tan
When landing on the slope,
tvgttv )cos(tan21)sin( 0
20
cossincoscossin2 0
gvt
cos)sin(2 0
gv
tvx )cos( 0
cos
)sin(cos2 20
gv
xy tan
2
20
cos)sin(cossin2
gv
cosxD
2
20
cos)sin(cos2
gv
How to find maximum D?
5 3( ) 15y x x x 2 25 ( 9) 0dy x x
dx
22
2 10 (2 9)d dy d y x xdx dx dx
How to find the max. or min. of a function ? *
Extreme value at 0dydx
;
2
2 0d ydx
2
2 0d ydx
2
2 0d ydx
Min.
Max.
Saddle point
0, 3x
12
x
Example : 3 2( ) 2 3 12 5y x x x x
Find the max. and min. of y(x)
2
212 6d dy d y x
dx dx dx
6( 2)( 1) 0x x 0dydx
1x 2
2 18 0d ydx
min 2 3 12 5 2y
2x 2
218 0d y
dx
max 16 12 24 5 25y
cosxD
20
2
2 cos sin( )cos
vg
What is the maximum D?20
2( ) 2 cos sin( )( )
cosdD d v
d d g
20
22 sin sin( ) cos cos( )( )
cosvg
20
22 cos( 2 )( )
cosvg
( ) 0, cos(2 ) 0,dDFord
2 ,2 ,
4 2
20
max 2
cos( ) sin( )2 4 2 4 2cos
vDg
20
22 1 sin( )
2cosvg
cosxD
2
20
cos)sin(cos2
gv
20
2( ) 2 cos( 2 )
cosdD v
d g
2 20
2 2( ) 4 sin( 2 )
cosd D v
d g
,4 2
2 20
2 2
sin( )( ) 4 2 0cos
d D vd g
' - r r u t
Origin O and O’coincide at t = 0, S’moves at u relative to S
dttud
dtrd
dtrd )(
uvv
dtud
dtvd
dtvd
aa
0
u
vv
Relative Velocity
Relative Velocity for 1D motion
MCv
MG MC CGv v v
M : Man ; C : Car ; G : Ground
CGv
ABv : Velocity of A relative to B
If vMC = 5 m/s and vCG = 3 m/s, then vMG = 8 m/s .
If vMC = 5 m/s and vCG = -3 m/s, then vMG = 2 m/s .
1. Motion in one dimension
Example 3-20 Car velocities at 900 . What is the velocity of A relative to B ( vAB ) ?
A
B
ˆ40.0( / )AGv m s i
ˆ40.0( / )BGv m s j
AB AG GB AG BGv v v v v
ˆ ˆ40.0( / ) 40.0( / )ABv m s i m s j
AGv
BGv ABv
2 240 40 ( / ) 56.6( / )ABv m s m s
vA =10m/s, vB =18m/s, D=36m, = 370 ,u=?
A Av0( 37 )
B Bv
D
u 210 /g m s
vA =10m/s, vB =18m/s, D=36m, = 370 ,u=? for hitting B
A Av0( 37 )
B Bv
ˆ ˆcos sinCAv u i u j
ˆ ˆ( )BA B A BAv v v i v i
; 0BA BA BAx D v t y
12 sin0CAuy t
g
2 2 sin2 cos sin 0BAvu u Dg g
;
D
u
1 1cosBAD v t u t
2 10 375 0u u 25( / )u m s
210 /g m sC
When C hits B, xCA = xBA and yCA = yBA
;cos tuxCA 2
21sin gttuyCA
sin 37 3 / 5cos 37 4 / 5
BG BW WGv v v
PG PA AGv v v
P : Plane ; A : Air ; G : Ground
B : Boat ; W : Water ; G : Ground
Real-Life Occasions for Relative Coordinate Systems
Example 4.10 A boat heading due north crosses a wide river with a speed of 10.0 km/h relative to the water. The water in the river has a uniform speed of 5.00 km/h due east relative to the Earth. Determine the velocity of the boat relative to an observer standing on either bank.
bE br rEv v v
2 2
2 2(10.0) (5.0) /11.2 /
bE br rEv v v
km hkm h
1tan ( ) 26.6rE
br
vv
brv
rEv
bEv
TEST 3-1
A
B
060
wind
The distance between A and B is 500 Km. The speed of the airplane relative to air is 500km/h.The speed of wind relative to earth is 40 Km/h.
What should it heading be ?
How long it take ?
Solution : sin sinA Ba b
0sin 40sin120sin 0.069500
a BAb
0 04 ; 56A C
02
0
sin 40sin 56 475.4 /sin sin 4
b Cc m sB
500 1.05475.4 /
Kmt hKm h
Bwind
C
A 060
A
CB a
bc
Solving Problem wih Newton’s Laws :Free Body Diagrams**
Chapter 4 The Law of Motion
Newton’s Laws
Copyright © 2012 Pearson Education Inc.
What are some properties of a force? - Figure 4.1
Contact forces Field forcesP. 84
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Drawing force vectors - Figure 4.3
• Use a vector arrow to indicate the magnitude and direction of the force.
Copyright © 2012 Pearson Education Inc.
Superposition of forces - Figure 4.4
• Several forces acting at a point on an object have the same effect as their vector sum acting at the same point.
Copyright © 2012 Pearson Education Inc.
Decomposing a force into its component vectors• Choose perpendicular x and y axes.
• Fx and Fy are the components of a force along these axes.
• Use trigonometry to find these force components.
Figure 4.5 :
Copyright © 2012 Pearson Education Inc.
Copyright © 2012 Pearson Education Inc.
Figure 4.7
In genearl,
Newton’s Laws of Motion
First law : 0 F
Second law:
AB BAF F
Third law:
The tendency of an object to resist any attempt to change its velocity is called inertia.
constant v
amF
