Chapter 4 – 2D and 3D Motion
I.
Definitions
II. Projectile motion
III. Uniform circular motion
IV. Non-uniform circular motion
V. Relative motion
I. Definitions
Position vector: extends from the origin of a coordinate system to the particle.

r  xiˆ  yˆj  zkˆ
(4.1)
Displacement vector: represents a particle’s position change during a certain
time interval.
  
r  r2  r1  ( x2  x1 )iˆ  ( y2  y1 ) ˆj  ( z 2  z1 )kˆ
Average velocity:


r x ˆ y ˆ z ˆ
vavg 

i
j
k
t t
t
t
(4.3)
( 4 .2 )
Instantaneous velocity:


d
r
dx ˆ dy ˆ dz ˆ
ˆ
ˆ
ˆ
 i
j k
v  vxi  v y j  vz k 
dt
dt dt
dt
(4.4)
-The direction of the instantaneous velocity of a
particle is always tangent to the particle’s path at
the particle’s position
 

v2  v1 v


t
t
(4.5)

dv y

dv
d
v
ˆj  dv z kˆ
a  a x iˆ  a y ˆj  a z kˆ 
 x iˆ 
dt
dt
dt
dt
(4.6)
Average acceleration: 
aavg
Instantaneous acceleration:
II. Projectile motion
Motion of a particle launched with initial velocity, v0 and free fall acceleration
g.
The horizontal and vertical motions are independent from each other.
- Horizontal motion: ax=0  vx=v0x= constant
x  x0  v0 x t  (v0 cos  0 )t
(4.7)
Range (R): horizontal distance traveled by a
projectile before returning to launch height.
- Vertical motion: ay= -g = constant
y  y0  v0 y t 
1 2
1
gt  (v0 sin  0 )t  gt 2
2
2
v y  v0 sin  0  gt
(4.9)
(4.8)
v y 2  (v0 sin  0 ) 2  2 g ( y  y0 )
(4.10)
- Trajectory: projectile’s path.
x0  y0  0
1 
x
x
x
( 4 .7 )  ( 4 .8 )  t 
g 
 y  v 0 sin  0

2  v 0 cos  0
v 0 cos  0
v 0 cos  0
gx 2
y  (tan  0 ) x 
2 ( v 0 cos  0 ) 2



2

( 4 . 11 )
- Horizontal range: R = x-x0; y-y0=0.
R  (v0 cos  0 )t  t 
R
v0 cos  0
2

1 2
1 
1
R
R
R2
  R tan  0  g 2
0  (v0 sin  0 )t  gt  (v0 sin  0 )

 g 
2 v0 cos 2  0
2
v0 cos  0 2  v0 cos  0 
2 sin  0 cos  0 2 v02
R
v0  sin 2 0
g
g
(4.12) (Maximum for a launch angle of 45º )
Overall assumption: the air through which the projectile moves has no effect
on its motion  friction neglected.
122: A third baseman wishes to throw to first base, 127 feet distant. His best throwing speed is 85 mi/h. (a) If
he throws the ball horizontally 3 ft above the ground, how far from first base will it hit the ground? (b) From
the same initial height, at what upward angle must the third baseman throw the ball if the first baseman is to
catch it 3 ft above the ground? (c) What will be the time of flight in that case?
y
 mi   1h   1609m 
 85   

  38m / s
h   3600 s   1mi 

3 feet    0.305m   0.91m
 1 foot 
v
0
h=3ft
B1
B3
xmax
0
xB1=38.7m
x
Horizontal movement
Vertical movement
xmax  x0  v0 x t
1
y  y0  v0 y t  gt 2
2
0  0.91m  4.9t 2  t  0.43s
xmax  0  38t  (38m / s )(0.43s )  16.4m from B3
The ball will hit ground at 22.3 m from B1
1 2
38 sin 
gt  v0 y  4.9t  v0 sin   t 
2
4.9
38.7 m
38.7
 v0 cos   t 
 1s
v0 x 
38 cos 
t
38.7
38 sin 

 189.63  1444 sin  cos  
38 cos 
4.9
0.13  0.5 sin 2    7.6
y  y0  0  v0 y t 
y
v
0
θ
h=3ft
B3
38.7m
B1
x
N7: In Galileo’s Two New Sciences, the author states that “for elevations (angles of projection) which exceed
or fall short of 45º by equal amounts, the ranges are equal…” Prove this statement.
y
  45
1  45  
v
v02
Range : R  sin 2 0  d max at h  0
g
 2  45  
0
θ=45º
x=R=R’?
x








v02
v02

R  sin 2 45    sin 90  2
g
g
v02
v02

R'  sin 2 45    sin 90  2
g
g
sin(a  b)  sin a cos b  cos a sin b
sin(a  b)  sin a cos b  cos a sin b




v02
v02


R
sin 90 cos(2 )  cos 90 sin(2 )  cos(2 )
g
g
v02
v02


R' 
sin 90 cos(2 )  cos 90 sin(2 )  cos(2 )
g
g
III. Uniform circular motion
Motion around a circle at constant speed.
Magnitude of velocity and acceleration constant.
Direction varies continuously.
-Velocity: tangent to circle in the direction of motion.
v2
a
r
- Acceleration: centripetal
- Period of revolution: T  2 r
v
vy
vx
(4.13)
(4.14)
  v  y p ˆ  v  x p  ˆ

 j
i  
v  v x iˆ  v y ˆj  (v sin  )iˆ  (v cos  ) ˆj  
r
r


 

   v2

 dv   v dy p  ˆ  v dx p  ˆ   v  ˆ  v  ˆ   v 2
ˆ



 ˆj




a
i
j
v
i
v
j
i

cos


sin









y
x




dt  r dt   r dt 
r
r
r
r


 

 

v2
v2
2
2
a

cos   sin  
r
r
a y sin 

a directed along radius  tan  

 tan 
a x cos 
a x2
 a 2y
54. A cat rides a merry-go-round while turning with uniform circular motion. At time t1= 2s, the cat’s velocity
is: v1= (3m/s)i+(4m/s)j, measured on an horizontal xy coordinate system. At time t2=5s its velocity is:
v2= (-3m/s)i+(-4m/s)j. What are (a) the magnitude of the cat’s centripetal acceleration and (b) the cat’s
average acceleration during the time interval t2-t1?
v
2
x
In 3s the velocity is reversed  the cat reaches the opposite
side of the circle
v1
y
v  32  4 2  5m / s
r
2r
T
 3s 
 r  4.77 m
v
5m / s
v 2 25m 2 / s 2
ac 

 5.23m / s 2
r
4.77 m
 

v2  v1 (6m / s )iˆ  (8m / s ) ˆj

 (2m / s 2 )iˆ  (2.67 m / s 2 ) ˆj
aavg 
t
3s

aavg  3.33m / s 2
IV. Non-Uniform circular motion
- A particle moves with varying speed in a circular path.
- The acceleration has two components:
- Radial  ar = -ac= -v2/r
- Tangential  at = dv/dt
- at causes the change in the speed of the particle.
a  ar2  at2

   d v ˆ v2
a  at  ar 
  rˆ
dt
r
- In uniform circular motion, v = constant  at = 0  a = ar
V. Relative motion
Particle’s velocity depends on reference frame
vPA  vPB  vBA
(4.15)
1D
Frame moves at constant velocity
0
d
d
d
(vPA )  (vPB )  (vBA )  a PA  a PB
dt
dt
dt
(4.16)
Observers on different frames of reference measure the same acceleration
for a moving particle if their relative velocity is constant.
75. A sled moves in the negative x direction at speed vs while a ball of ice is shot from the sled with a velocity
v0= v0xi+ v0yj relative to the sled. When the ball lands, its horizontal displacement ∆xbg relative to the ground
(from its launch position to its landing position) is measured. The figure gives ∆xbg as a function of vs. Assume
it lands at approximately its launch height. What are the values of (a) v0x and (b) v0y?
The ball’s displacement ∆xbs relative to the sled can also be measured. Assume that the sled’s velocity is not
changed when the ball is shot. What is ∆xbs when vs is (c) 5m/s and (d) 15m/s?

vs  vs iˆ

v0,rel  v0 x iˆ  v0 y ˆj

v0 g  (v0 x  vs )iˆ  v0 y ˆj
Launch velocity relative
to ground
∆xbg= a+ b vs
Displacements relative to ground
xland  xlaunch  xbg  (v0 x  vs )t flight
yland  ylaunch  0 
2v0 y
v0 y t flight  0.5 gt 2flight
80m
 v0 y  19.6m / s
20m / s
g
2v
2  (10m / s )
xbg  0  vs  10m / s  0  0 x 
 v0 x  10m / s
g
g

2v0 y
0  v0 y t flight  0.5 gt 2flight  t 
g
b
a
2v0 y 2v0 x v0 y 2v0 y
xbg  (v0 x  vs )


vs
g
g
g

Displacements relative to the sled
xbs  v0 x t flight  (10m / s )  4s  40m
t flight 
2v0 y
g
 4s
v0  10 2  19.6 2  22m / s
Relative to the sled, the displacement does not depend
on the sled’s speed Answer (c)= Answer (d)
(iii) A dog wishes to cross a river to a point directly opposite as shown. It can swim at
2m/s in still water and the river is flowing at 1m/s. At what angle with respect to the
line joining the starting and finishing points should it start swimming?

vr  (1m / s )iˆ
y
finish

v0  2 sin  iˆ  2 cos  ˆj

vr
1m/s

vf
θ
start

v0

 
v f  vr  v0  ( iˆ  2 sin  iˆ  2 cos  ˆj ) m / s  (v f ˆj ) m / s
x
2 sin   1  0    30 
2 cos   v f  v f  3m / s
(ii) A particle moves with constant speed around the circle below. When it is at
point A its coordinates are x=0, y=3m and its velocity is (5m/s)i. What are its
velocity and acceleration at point B? Express your answer in terms of unit
vectors.
y

v 2 ˆ 25m 2 / s 2 ˆ
aB  i 
i  (8.3m / s 2 )iˆ
3m
r
A

v A  (5m / s )iˆ

vB  (5m / s ) ˆj
B

a B  (8.3m / s 2 )iˆ
x
120. A hang glider is 7.5 m above ground level with a velocity of 8m/s at an angle of 30º below the
horizontal and a constant acceleration of 1m/s2, up. (a) Assume t=0 at the instant just described and write
an equation for the elevation y of the hang glider as a function of t, with y=0 at ground level. (b) Use the
equation to determine the value of t when y=0. (c) Explain why there are two solutions to part B. Which one
represents the time it takes the hang glider to reach ground level? (d) how far does the hang glider travel
horizontally during the interval between t=0 and the time it reaches the ground? For the same initial position
and velocity, what constant acceleration will cause the hang glider to reach ground level with zero velocity?
Express your answer in terms of unit vectors.
y
1
y  y0  v0 y t  at 2  y  7.5  4t  0.5t 2
2
y  0  0  7.5  4t  0.5t 2  t1  5s, t 2  3s
30º
v0= 8m/s
h=7.5m

a  1 ˆj

vo  8 cos 30 iˆ  8 sin 30 ˆj
y(m)
x
0
7.5
0
1 2 3 4
t(s)
5
If the ground was not solid, the glider would swoop down,
passing through the surface, then back up again, with the two
times of passing being t=3s, t=5s.
d max  v0 x t  (6.93m / s )  (3s )  20.78m


y  0  v f  v fx iˆ  v fy ˆj  0
Vertical movement :
v y 2  v02 y  2a y  ( y  y0 )  4 2  15a y  a y  1.1m / s 2
v y  v0 y  a y t  0  0  4  1.1t  t  3.75s
Horizontal movement :
0  v x  v0 x  a x  t  6.93  3.75a x  a x  1.85m / s 2

a  (1.85m / s 2 )iˆ  (1.1m / s 2 ) ˆj
40. A ball is to be shot from level ground with certain speed. The figure below shows the range R it will
have versus the launch angle θ0 at which it can be launched. The choice of θ0 determines the flight time; let
tmax represent the maximum flight time. What is the least speed the ball will have during its flight if θ0 is
chosen such as that the flight time is 0.5tmax?
R(m)
200
y  y0  0  v0 y t  0.5 gt 2  v0 sin  0t  4.9t 2
t 
t flight 
2v0 sin  0
2v
 sin  0  1  t max  0
g
g
100
2v0 sin  0
2v sin  0 v0
 0.5t max  0
   0  sin 1 0.5  30
g
g
g
The lowest speed occurs at the top of the trajectory (half of total time of
flight), when the velocity has simply an x-component.
vmin  v x at half trajectory  v0 x  v0 cos 30 for 0.5t max
Graph Rmax  240m for  0  45
v0 2
v0 2
v0 2

R
sin 2 0  240m 
sin 90 
 v0  48.5m / s
g
g
g
vmin  v x  (48.5m / s ) cos 30  42m / s
θ0