# CH4s Motion in Two Dimensions ```Chapter 4
Motion in Two Dimensions
Position Vector
r  r f − ri
Velocity Vector
v avg
r

t
r d r
v  lim
=
t → 0 t
dt
v= v
Acceleration Vector
a avg 
v dv
=
t → 0 t
dt
a  lim
v v f − v i
=
t
t f − ti
Two-Dimensional Motion with Constant Acceleration
r = xˆi + yˆj
Two-Dimensional Motion with Constant Acceleration
v=
dr dx ˆ dy ˆ
=
i+
j = vx ˆi + v y ˆj
dt dt
dt
vxf = vxi + ax t and v yf = v yi + a y t
(
) (
)
v f = ( vxi + ax t ) ˆi + ( v yi + a y t ) ˆj = vxi ˆi + v yi ˆj + ax ˆi + a y ˆj t
v f = v i + at ( for constant a )
Two-Dimensional Motion with Constant Acceleration
1
1
x f = xi + vxi t + ax t 2 y f = yi + v yi t + a y t 2
2
2
1
1

 

r f =  xi + vxi t + ax t 2  ˆi +  yi + v yi t + a y t 2  ˆj
2
2

 

1
= xi ˆi + yi ˆj + vxi ˆi + v yi ˆj t + ax ˆi + a y ˆj t 2
2
(
) (
1
r f = ri + v i t + at 2
2
)
(
( for constant a )
)
Two-Dimensional Motion with Constant Acceleration
v f = v i + at


1

r f = ri + v i t + at 2 
2 
( for constant a )
Example 4.1: Motion in a Plane
A particle moves in the xy plane, starting from the origin at t = 0 with an initial
velocity having an x component of 20 m/s and a y component of −15 m/s. The
particle experiences an acceleration in the x direction, given by ax = 4.0 m/s2.
(A) Determine the total velocity vector at any later time.
Example 4.1: Motion in a Plane
Example 4.1: Motion in a Plane
(B) Calculate the velocity and speed of the particle at t = 5.0 s and the angle
the velocity vector makes with the x axis.
Example 4.1: Motion in a Plane
(C) Determine the x and y coordinates of the particle at any time t and its
position vector at this time.
Projectile Motion
Projectile Motion
1
r f = ri + v i t + gt 2
2
vxi = vi cos i
v yi = vi sin i
Projectile Motion
x f = xi + vxi t
v yf = v yi − gt
v y ,avg =
v yi + v yf
2
1
y f = yi + ( v yi + v yf ) t
2
1
y f = yi + v yi t − gt 2
2
v yf 2 = v yi 2 − 2 g ( y f − yi )
Horizontal Range and Maximum Height of a Projectile
Height of a Projectile
v yf = v yi − gt → 0 = vi sin  i − gt A → t A =
vi sin  i
y f = yi + v yi t −
g
1 2
gt →
2
v sin i 1  vi sin i 
h = ( vi sin i ) i
− g

g
2  g 
vi 2 sin 2 i
h=
2g
2
Horizontal Range of a Projectile
x f = xi + vxi t → R = vxi t B = ( vi cos  ) 2t A
2vi sin i 2vi 2 sin i cos i
= ( vi cos i )
=
g
g
2sin i cos i = sin 2i
vi 2 sin 2i
R=
g
Rmax
vi 2
=
g
Trajectories of a Projectile
Example 4.2: The Long Jump
A long jumper leaves the ground at an angle of 20.0&deg; above the horizontal and
at a speed of 11.0 m/s.
(A) How far does he jump in the horizontal direction?
Example 4.2: The Long Jump
(B) What is the maximum height reached?
Example 4.3: A Bull’s-Eye Every Time
In a popular lecture demonstration, a
projectile is aimed directly at a target in
such a way that the projectile leaves the
gun at the same time the target is dropped
from rest. Show that the projectile hits the
falling target as shown in the figure.
Example 4.3: A Bull’s-Eye Every Time
Example 4.4: That’s Quite an Arm
A stone is thrown from the top of a building upward at an angle of 30.0&deg;
to the horizontal with an initial speed of
20.0 m/s as shown in the figure. The
height from which the stone is thrown is
45.0 m above the ground.
(A) How long does it take the stone to
reach the ground?
Example 4.4: That’s Quite an Arm
Example 4.4: That’s Quite an Arm
(B) What is the speed of the stone just before it strikes the ground?
Example 4.5: The End of the Ski Jump
A ski jumper leaves the ski track moving in the horizontal direction with a
speed of 25.0 m/s as shown in the figure. The landing incline below her falls
off with a slope of 35.0&deg;. Where does she land on the incline?
Example 4.5: The End of the Ski Jump
Analysis Model: Particle in Uniform Circular Motion
dv
a=
dt
Analysis Model: Particle in Uniform Circular Motion
v
v
=
r
r
a avg =
v
t
=
v r
r t
Centripetal Acceleration
v2
ac =
r
Particle in Uniform Circular Motion
v=
2 r
2 r
→ T=
v
T
2
=
T
 v  v
 = 2 
 = → v = r
 2 r  r
ac =
( r )
r
2
= r 2
Analysis Model: Particle in Uniform Circular Motion
v2
ac =
r
2 r
T=
v
2
=
T
Example 4.6: The Centripetal Acceleration of the Earth
(A) What is the centripetal acceleration of the Earth as it moves in its orbit
around the Sun?
Example 4.6: The Centripetal Acceleration of the Earth
(B) What is the angular speed of the Earth in its orbit around the Sun?
a = a r + at
dv
at =
dt
v2
ar = −ac = −
r
a = ar 2 + at 2
Example 4.7: Over the Rise
A car leaves a stop sign and exhibits a constant acceleration of 0.300 m/s2
parallel to the roadway. The car passes over a rise in the roadway such that
the top of the rise is shaped like an arc of a circle of radius 500 m. At the
moment the car is at the top of the rise, its velocity vector is horizontal and
has a magnitude of 6.00 m/s. What are the magnitude and direction of the
total acceleration vector for the car at this instant?
Example 4.7: Over the Rise
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