2D - 1
Motion in 2D
Velocity and acceleration are vectors. They can have any direction. When we are considering
motion in the xy plane, these vectors can point anywhere in the plane. A common example of
motion in 2D is...
Projectile motion
Consider a projectile fired from a cannon, with an initial velocity v 0 with a direction of  above
the horizontal.
y
v 0x  v 0 cos 
vy
vo
v 0 y  v 0 sin 

x
vx
Acceleration is a vector, and can have any direction. But in the special case of acceleration due
solely to gravity, the acceleration is always straight down.
y
a
ax = 0
ay = –g
x
Review of 1D motion:
dv
dx
a 
, v 
,
dt
dt
From these two equations, we can derive, for the special case a = constant,
(a)
v = vo + a t
(b)
x = x o + v o t + (1/ 2) a t 2
(c)
v 2 = v o 2 + 2 a (x - x o )
vo + v
2
xo , vo = initial position, initial velocity
(d)
v =
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x, v = position, velocity at time t
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2D - 2
Suppose that a = 0 . In this case v = constant, and v  v 
 v0 
x  x0
,
t
v0 t  x  x 0 ,
x  x 0  v0 t
If a ≠ 0 then

position if a = 0
x
 v 0  constant
t
x  x 0  v0 t
1
2
a t2
how much more (a > 0)
or less (a < 0) distance you
go if a  0
End of 1D motion review.
Now, 2D Motion
a
dv
dt
and
v

ax 
dr
dt
d vy
d vx
, ay 
dt
dt
 vx 
Special case: a  constant
dx
dy
, vy 
dt
dt
 a x  constant, a y  constant
This is exactly like the 1D motion case, except now we have separate equations for x-motion and
y-motion. We can treat the x-motion and y-motion separately:
x = x o + v ox t +
1
2
ax t2
ü
ïï
ý X
ïï
þ
1
2
ay t2
ü
ïï
ý Y
ïï
þ
v x = v ox + a x t
y = y o + v oy t +
v y = v oy + a y t
These are the x- and y-components of the vector
equations
v
v
v
v
r = ro + v o t + 12 a t 2
v
v
v
v = vo + a t
Example: Horizontal Rifle. A rifle bullet is fired horizontally with vo = 100 m/s from an initial
height of y0 = 2.0 m. Assume no air resistance. How long is the bullet in flight? How far does
the bullet go before it hits the ground?
y
vox = 100 m/s
2m
x
Key idea in all projectile motion problems: treat x- and y-motions separately!
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2D - 3
The motion along the y-direction (vertical motion) is completely independent of the motion
along the x-direction (horizontal motion).
X: x 0  0
Y: y 0  2 m
vox  100 m / s
v oy  0
ax  0
a y   g   9.8 m / s 2
The time to hit the ground is entirely controlled by the y-motion:
v oy t {y = y o + {
0
1
2
g t2
Þ
0 = y0 -
1
2
g t2 ,
y0 =
1
2
g t2
0
2 y0 = g t 2 , t 2 =
2 y0
,
g
t =
2 y0
g
=
2 (2)
= 0.64s
9.8
Now we look at the x-equations to see how far along the x-direction the bullet traveled in 0.64 s.
ax = 0 Þ
x =
v x = constant = vox = 100 m / s
x + vox t +
{o
1
2
a x t 2 = v ox t = 100(0.64) = 64 m
{
0
0
Why vx = constant ? The force of gravity is straight down. There is no sideways force to change
vx (assuming no air resistance).

Another question: What is the speed of the bullet as it falls?
v x  constant  v o x  v o
v y  v oy  a y t   g t
0
g
As the bullet travels, its vx remains constant, while |vy | grows larger and larger.
y
vox = vo
vx
vy
v
x
vx
v
2/6/2016
|vy |
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2D - 4
speed = magnitude of velocity = v

vx 2  v y 2
vox 2   g t 

2
The speed is a minimum at t = 0 when vy = 0 (the moment when the bullet leaves the gun). The
speed is maximum when vy is maximum, just before the bullet reaches the ground. Don’t forget
that we are assuming no air resistance. For a real rifle fired in real air, the bullet’s speed is
usually maximum when leaving the barrel, and then air resistance slows the bullet down as it
travels.
Example: A projectile is fired on an airless world with initial speed vo at an angle  above the
horizontal. What is the minimum speed of the projectile? Answer: vox = vo cos .
a x  0  v x  constant = vox  vo cos 
Proof:
a y   g  v y  voy  g t
y
vy
( vy = 0 )
vox
vox
voy

vox
x
Here, the speed is minimum at the top of the trajectory, where vy = 0.
Review of acceleration:
1D: a 
dv
dt
v  v2  v1 means
2D: a 
dv
v

dt
t
v2  v1  v


v is the vector you add to v1 to get v 2 .
v 2  v1
t
v1
v
v2
The direction of a is the same as the direction of v
(since a  v  positive number ( 1/ t )
V1
V1
V2
V2
x
The direction of the acceleration of gravity is the direction of V: straight down!
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V
2D - 5
"Shoot the Monkey" Experiment:
A hunter aims a rifle at a monkey hanging in a tree. The rifle fires at the same instant that the
monkey lets go and drops. Does the bullet hit the poor monkey? Answer: Yes!
First, consider the situation with no gravity :
If no gravity, the bullet goes in a straight
line, and the monkey does not fall. So the
monkey is hit.
y
The height of the bullet (with ay = 0) is

y

x
y0  v0y t

v0 sin  t
0
Now, turn on gravity. The height of the bullet is now: y
 y0  v0y t 
1
2
g t2 .
y
(1/2)g t2
x
With gravity on, the bullet falls below the straight-line path by a distance (1/2)g t2 , which is
exactly the same distance that the monkey falls. So the monkey falls into the path of the bullet.
Poor monkey!
Circular Motion and Acceleration
Circular motion: consider an object moving in a circle of
radius r, with constant speed v.
T = period = time for 1 complete revolution, 1 cycle
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r
v
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2D - 6
speed v  v 
2 r
distance

time
T
An object moving in a circle is accelerating, because its velocity is changing -- changing
direction. Recall the definition of acceleration:
a 
dv
dt

v
v  v1
, velocity v can change is two ways:
 2
t
t
Magnitude can change or
direction can change:
v
v
v1
v2
v2
v1
For circular motion with constant speed, we will show that
v2
1) the magnitude of the acceleration is a  a 
r
2) the direction of the acceleration is always towards the center of
motion. This is centripetal acceleration. "centripetal" = "toward center"
Notice that the direction of acceleration vector is always changing,
therefore this is not a case of constant acceleration (so we cannot use
the "constant acceleration formulas")
r
v
a
a
v
Is claim (1) sensible?
 v2 
Check units:   
 r 
m s
m
2

 m2 
 s2 


m

m
s2

a 
Yep.
Think: to get a big a, we must have a rapidly changing velocity. Here, we need to rapidly change
the direction of vector v  need to get around circle quickly  need either large speed v or a
small radius r.  a = v2 / r makes sense. (Proof is given below.)
Is claim (2) sensible?
Observe that vector v is toward center of circle.
v1
v2
v
Direction of a = direction toward which
velocity is changing
v1
v2
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2D - 7
Example: acceleration on a merry-go-round. Radius r = 5 m , period T = 3 s
v
2 r
T

2 (5)
 10. 5 m/s
3
r
v
a
a =
v2
10.5

r
5
1g
2
 22 m/s 2 
9.8 m / s 2
 2.3 g's !
A human can withstand an acceleration of about 5 g's for a few minutes or ~10 g's for a few
seconds without losing consciousness.
Proof of a = v2 / r for circular motion with constant speed
The proof involves geometry (similar triangles). It is
mathematically simple, but subtle.
Consider the motion of a particle on a circle of radius r with
constant speed v. And consider the position of the particle at two
times separately by a short time interval t. (In the end we will
take the limit as t  0.) We can draw a vector diagrams
representing r1  r  r2 and v1  v  v2 :
r2
r1
v
r2
r = v t
r1
v2
v1
Notice that these are similar triangles (same angles, same length ratios). Also, note that
r1  r2  r and v1  v2  v .
v t
v
r


. A little algebra gives
v
r
r
v v
v
v2

. Finally, we take the limit t  0 and get acceleration a 
.
t
r
r
Because the triangles are similar, we can write
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v2
v1