Download 4.1 - 4.3 - OoCities

4 Tow-Dimensional Kinematics
4-1 Motion in Two Dimensions
Recall the equation for motion in 1dimension
vf=vi+at
xf=xi+1/2(vi+vf)t
Δx=xf - xi
Δx=vit+1/2at2
vav=(vi+vf)/2
vf2=vi2+2aΔx
In 2 dimension apply the equation of
motion to x & y directions

ri ( xi , yi )
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

ri  xi x  yi y

rf ( x f , y f )

vi (vix , viy )



rf  x f x  y f y



vi  vix x  viy y

v f (v fx , v fx )


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v f  v fx x  v fy y

a (a x , a y )
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
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a  ax x  a y y
Equation of motion along x; Equation of motion along y
vfx=vix+ax t
vfy=viy+ay t
xf=xix+1/2(vix+vfx)t
yf=xiy+1/2(viy+vfy)t
Δx=vixt+1/2ax t2
vfx2=vix2+2ax Δx
Δy=viyt+1/2ay t2
vfy2=viy2+2ay Δy
same time t !
Constant Velocity
ax=0
vfx=vix=vx
Δx=vxt
ay=0
vfy=viy=vy,
Δy=vyt
Ex1.
Ball velocity is 2 m/s, in a direction 300 with
horizon, the ball travel 3 m along the x, find
the displacement in y.
Constant Acceleration
Ex. 4-2 Hummer Acceleration
A hummingbird is flying in such a way that it
is initially moving vertically with a speed of
4.6 m/s and accelerating horizontally at 11
m/s2 . Assuming the bird’s acceleration
remains constant for the time interval of
interest, find the horizontally and vertical
distance through which it moves in 0.55 s.
4-2 Projectile Motion: Basic
Equations
Projectile motion, assumptions:
• Air resistance ignored
• The acceleration due to gravity is constant,
downward, and has a magnitude equal to g=9.81 m/s2
• The earth’s rotation is ignored.
Projectile motion
x-direction
y-direction
ax=0
ay=-g=-9.8 m/s2
xf=xi+vixt
vyf=viy-gt
vyf2=viy2-2gΔy
4-3 Zero Launch Angle
constant velocity along x direction
• Sketch
• Choose coordinate system
• Known & unknown quantities in x & y
direction
• Apply equations
The ball from a height 100 m with the 10 m/s
horizontal velocity throws out, find how long
the ball touch the ground and the horizontal
displacement.
y
Vix=10m/s
h=100 m
xi
xf
x
Jumping a crevasse
From 2.75 m high, and jump 4.10 m the width
of crevasse, find the minimal speed to land
the other side. Find the land speed.
vo
2.75 m
4.10 m
• Parabolic Path
x=vixt
yf =yi+1/2ayt2=h+1/2(-g)t2= h-1/2gt2=h-(g/2vix2)x2
t=x/vix
y=a+bx2