2D and 3D Kinematics!
General comment: Displacement, velocity and acceleration
in 2D and 3D very similar to 1D, but the use of vector
relationships will be exercised to the fullest extent.!
Key: Break down into components, now there are 2 or
3 of them, treat them independently, but recognize the
general link between them. !
Example: a ball thrown horizontally with a speed v,
continues to move horizontally with speed v even as it
falls with an increasing speed in the vertical direction
(due to gravity). Common link: time!
Common problems: projectile motion, hitting baseballs,
boat swayed by current, circular motion !
!"
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Position Vector!
Displacement Vector!
Average Velocity!
(two vectors are parallel since
t is a scalar)!
Can write as!
(Instantaneous) Velocity!
2"
! dx
dy ˆ dz ˆ
v = iˆ +
j+
k
dt
dt
dt
Acceleration !
!
!
dv d 2 r
a=
=
dt dt 2
Similar to velocity, 2D and 3D also have multiple
components !! dv
dv y
dv z
!
!
a=
x
dt
iˆ +
dt
ˆj +
dt
kˆ
In terms of displacement (in components)!
!
! d 2 x ˆ d 2 y ˆ d 2z ˆ
a= 2 i + 2 j+ 2 k
dt
dt
dt
Remark: Similar to 1D problems, you CAN use constant
acceleration equations ONLY IF acceleration (2D, 3D) is
constant!!
3"
!
Simple Vector Calculations (constant velocity)!
An owl perched on a tree 19.5 m above the a mouse
standing still. The owl pushes off from the branch and
descends towards the mouse. By adjusting its body in
flight, the owl maintains a constant speed of 3.1 m/s at
an angle of 20o below horizontal.!
(a)! How long will it take the owl to land on the mouse?!
(b)! How far has the owl traveled in the horizontal
direction when it landed. !
Solution: !
y0=19.5 m, !
v=constant, !
Angle=-20o, !
(a)! t = ? !
Draw a diagram!
!"
!=20"
!4536!"17/"
!869"1"
:"
Break into components (note I used scalars to indicate
they are components along the corresponding vector
directions, ! v = v cos(20) = 3.1cos(20) = 2.91 m/s
0x
0
A fairly large value above means most of the velocity
component is in the x-direction. Speed in –Y direction,!
! v 0 y = v 0 sin(20) = 3.1sin(20) = 1.06 m/s
Constant speeds in both x and -y directions,!
y = y 0 " v 0 y t = 19.5 "1.06 t
0 = 19.5 "1.06 t
t = 18.44 s
!
Constant speeds in both x and y directions,!
! x = x 0 + v 0 x t = 0 + 2.91" 18.44 = 53.66 m
!
Remark:
!
!
!
(1)! Alternatively, due to constant velocity, we can calculate this by
simple geometry of
x= y/tan(20)!
(2) note I have used mainly scalar component for each direction! 9"
Example (variable velocities, 2D)!
Consider an aeroplane taking off. At the moment
the wheels leave the ground the plane is
travelling at 70m/s horizontally. The wings
generate a lift which causes a 1ms-2 vertical
acceleration and the jets cause a 2ms-2
horizontal acceleration. Neglecting air resistance
what is the horizontal distance of the plane from
the airport when it has climbed to 5000m and
what is its velocity?!
!!
Identify: a variable velocity problem in 2D, geometry of
vertical vs. horizontal distance (last problem) won’t work.!
Setup+ Execute:
First, draw a diagram, then write
known and unknowns!
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Example (variable velocities)!
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!!
!!
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Get time from vertical component, ! y = v t + 1 a t 2
0y
y
1
5000 = 0 + "1" t 2
2
2
t = 100 s
Now, time is a common link between components, we can
!
now compute horizontal distance (same equation),!
!
1
x = 70 "100 + " 2 "10000 = 17,000 m
2
! x! = v! t + 1 a! t 2
0x
x
2
Part 2:
!
!*"
final velocity calculation.!
Again, break into x and y components, vertically!
!v y = v 0y + ay t = 0 + 1"100 = 100 m/s
x components, v! x = v 0x + ax t = 70 + 2 "100 = 270 m/s
!"
vector op, !v! = v iˆ + v ˆj = 270iˆ + 100 ˆj
!"
x
y
! !G" magnitude, ! v! = 270 2 + 100 2 = 287.92 m/s
!
"1
!
Angle: ! arg(v ) = tan (100 /270) = 20.32 deg P"
!
!
Projectile Motion!
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!!
8"
Testing the understanding!
You and your little brother are doing a cannonball at a
pool. Obviously being bigger and faster you have
greater takeoff speed. Who will land in water first?
Who will have higher speed upon landing?!
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Can you think of situations where your answers would need modifying?
!!"
Also, which has higher speed?!
!
A special Case of Projectile Motion!
(zero launch angle)!
Lets look at the cannonball problem. You run toward
the pool at a constant speed of 4 m/s. The water level
is 1.8 m below the platform surface. How far away
from the edge of the pool did you land and what is
your landing speed and angle with horizontal? (ASSUME !
g = 10 m/s2)!
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Identify: This is a
!"
zero launch angle
problem. !
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Vertical = free fall,
G"
horizontal = constant v!
Setup: " Draw a diagram!
Set water depth at y=0 !
!2"
I need the time, but I can’t find from horizontal
direction since it involves t. So work with vertical,
1
y = v 0y t " gt 2
2
Now can find x!
!
t=
2y
=
g
!
3.6
= 0.6 s
10
x = v x t = 4 " 0.6 = 2.4 m
To find velocity, !
need both x and y components,!
X component,! v x = v 0x = v 0 = 4 m/s
!
y component,!
vector op, !
!
magnitude, !
!
Angle: !
!
v y = v 0y " gt = 0 "10 # 0.6 = 6 m/s
!
v = v x iˆ + v y ˆj = 4 iˆ + 6 ˆj
!
v = 4 2 + 6 2 = 7.35 m/s
!
arg(v ) = tan"1 (6 /4 ) = 56.31 deg
Evaluate:
!
the numbers are !
reasonable!
!3"
!
Homerun problem!
Slugger Jason Bay (Canadian with the Redsox) hits a
baseball with an initial speed v at an angle ! above the
horizontal. Neglecting air resistance, what angle ! gives
the maximum range of the ball, what is that range and
what is the maximum height reached?!
Solution:
Identify this as a projectile problem
Draw diagram!
!
-+
,)
,+
!)
+)
.+
!:"
Maximum Range!
Note: This is a problem working without numbers, !
treat K"and ! as given."
At start point A, vertical component: v!0y = v 0 " sin # = v sin #
Lets find the total time it takes the ball to land:"
2v sin "
1
t=
0 = v sin "t # gt 2
g
2
!
2v sin "
Plug into x direction:" x = v 0x t
x = v cos" #
g
2
2v 2 sin " cos"
v
sin2
"
x=
x=
!
g
!
g
1
y = v 0y t " gt 2
2
!
The maximum!of x is obtained if !"5":9&"
where x max
!
!v 2 "1 v 2
=
=
g
g
!
This means if Jason is “swing for the fences” (hit as far
!9"
as he can) he should make contact at 45 deg angle!"
!
Height of the ball for xmax case!
This involves points A and B, vertical component !
v 0y = v sin "
v h max = 0
ay = "g
h=?
Now use the vertical velocity/displacement equations!
!
!
!2 ! 2
v y = v 0y + 2ah
!
!
!
!
!2 ! 2
h = (v y " v 0y ) /2("g)
!
h = (0 " v 2 sin 2 # ) /2("g)
2
"
%
2
h = v 2 $ ' /2g
# 2 &
v2
h=
4g
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