CH 4 – Motion in Two Dimensions
Vector review
A vector has magnitude and direction.
Examples: displacement, velocity, force, magnetic field
A scalar has magnitude but no direction.
Examples: distance, speed, mass, temperature
Addition of vectors
R = A + B (= B + A)
A
R=A+B
B
B
C
R=A+B+C
R=B+A
B
A
A
Subtraction of vectors: R = A – B = A + (-B)
A
B
-B
Multiplication of a scalar times a vector:
C = aB
-B
R=A-B
B
Example: C = -3B
-3B
1
Components of a vector:
In rectangular coordinates, a vector can be resolved
into x- and y- components.
A  Ax  Ay
A
Ay
Ax  A cos  , Ay  A sin 
A
Ax  Ay
2
y

2
Ax
tan   Ay / Ax
x
Adding vectors using components:
R  AB
R x  Ax  B x
R y  Ay  B y
R  Rx 2  R y 2
tan   R y / R x
Example: A boy walks 100 m 45o north of east, then walks 75 m 30o south of east. What
is his net displacement (magnitude and direction)?
A  100 m @ 45 o
B  75 m @  30 o
Ax  100 m cos( 45 o )  70.7 m
y
(north)
Ay  100 m sin( 45 o )  70.7 m
30o
100 m
75 m
B x  75 m cos( 30 )  65.0 m
o
139.7 m
B y  75 m sin( 30 )  37.5 m
o
o
45
R x  70.7 m  65.0 m  135.7 m
R y  70.7 m  37.5 m  33.2 m
13.4o
x (east)
R  (135.7) 2  (33.2) 2  139.7 m
  tan 1 ( R y / R x )
 tan 1 (33.2 / 135.7)  13.4 o
2
Displacement, velocity, and acceleration in 2-D
Displacement: r  r f  ri
Average velocity: v 
r
t
y
r dr
Instantaneous velocity: v  lim

dt
t  0 t
r
Velocity is always in a direction tangent to the
path
v
Average acceleration: a 
t
v dv
Instantaneous acceleration: a  lim

dt
t  0 t
ri
rf
x
y
Projectile motion
x  v x 0 t
x-direction:
v0
v x  v x 0  const
v0y
0
y  v y 0 t  12 gt 2
v0x
y-direction: v y  v y 0  gt
vx 0  v0 cos  0
v y 2  v y 0 2  2 g y
total velocity:
x
v y 0  v0 sin  0
v  vx 2  v y 2
  tan 1 (v y / v x )
Example:
-
Time of flight of a projectile over flat ground.
y  v y 0 t  12 gt 2  0  t (v y 0  12 gt )  t  0 ( start ) or t 
-
Range of a projectile over flat ground.
x  v x 0 t 
2v x 0 v y 0
g
2v0 sin  0 cos  0
g
2

3
2v y0
g
(finish)
-
Maximum height of a projectile. At peak height
 v y0  1  v y0 
  2 g

v y  v y 0  gt  0, t 
, y  v y 0 t  gt  v y 0 

 g 
g
g




2
v y0
Or, y 
2g
v y0
1
2
2
2
Example: A football is kicked from the ground with an initial speed of 25 m/s at an angle
of 30o above the ground.
v x 0  v0 cos  0  (25 m / s ) cos(30 o )  21.7 m / s
v y 0  v0 sin  0  (25 m / s ) sin( 30 o )  12.5 m / s
2v y 0 2(12.5)
hang time 

 2.55 s
g
9. 8
range  v x 0 t  (21.7)( 2.55)  55.4 m
maximum elevation 
v y0 2
2g

(12.5) 2
 8.0 m
2(9.8)
What is the speed and direction of motion of the all 2 s after it is kicked?
v x  v x 0  21.7 m / s, v y  v y 0  gt  12.5  (9.8)( 2)  7.1 m / s
v  v x 2  v y 2  (21.7) 2   7.12  22.8 m / s
  tan 1 (v y / v x )  tan 1 (7.1 / 21.7)  18.1o
Example: A projectile is fired from the edge of the roof of a building 20 m tall with an
initial speed of 25 m/s at an angle of 30o above the horizon.
-
How long does it take for the projectile to reach the ground below?
y  v y 0 t  12 gt 2
 20  25(sin( 30 o )t  12 (9.8)t 2  12.5t  4.9t 2
4.9t 2  12.5t  20  0 (of form at 2  bt  c  0)
t
2
 b  b 2  4ac 12.5  12.5  4(4.9)( 20) 12.5  23.4


 3.66 s,  1.10 s
2a
2(4.9)
9.8
4
-
What is the speed of the projectile just before it lands?
v x  v x0  v0 cos  0  25 cos(30 o )  21.7 m / s
v y  v y 0  gt  25 sin( 30 o )  9.8(3.66)   23.4 m / s
v  (21.7) 2  (23.4) 2  31.9 m / s
For further thought:
-
How far does the projectile land from the base of the building?
What would be the final speed if the projectile were thrown straight down with a
speed of 25 m/s instead of being up at an angle of 30o?
What is the meaning of the negative time in the solution?
Can you solve for the final speed without using the quadratic formula?
Can you solve for the time of flight without using the quadratic formula?
Centripetal acceleration
An object moving in a circle at constant speed is said to undergo uniform circular motion.
Even though the speed is constant, its direction is changing – thus, it is accelerating. The
acceleration is directed toward the center of the circle and is referred to as radial or
centripetal acceleration. The magnitude of this acceleration is given by
v1
v2
.
ac 
r
v2
s

v1
v 
v2
r
The diagram to the right shows how this formula
is obtained for uniform circular motion.
Because of the change in direction, the velocity
changes by an amount v which is directed
toward the center of the circle. From the
similarity of the triangles formed by sides s
and r and by v and v, we have
s v

r
v
vs
v 
r
v = |v1| = |v2|
v
 
ac
ac
v
ac
v v s v
v2
ac 

 v
t r t r
r
v
5
Example:
A car goes around a circular track of circumference 2000 m with constant speed of 1
minute. What is the magnitude of the acceleration of the car?
v
s 2000 m

 33.3 m / s
t
60 s
r
s 2000 m

 318 m
2
2
ac 
v 2 (33.3 m / s ) 2

 3.5 m / s 2
r
318 m
Combined tangential and centripetal acceleration
An object can also have a component of acceleration tangent
to its path if its speed is changing. This component if given
by
at
v
a
ac
at 
dv
dt
If the object is also moving in a curved path as its speed is changing, then the total
acceleration is given by
a  ac  a t
Since these two components of the acceleration are mutually perpendicular, then the
magnitude of the total acceleration is
dv
v2
a  at  ac , where at 
and ac 
dt
r
2
2
Relative Velocity
The velocity of an object (A) depends on the coordinate system in which it is measured.
If two systems, B and C, are moving with respect to each other, then velocities transform
as
v AC  v AB  v BC
6
Example: A boat can travel with a speed of 4 m/s in still water.
-
If the boat heads straight across a river
flowing at 3 m/s, what is the speed of the
boat with respect to the earth?
VWE
VBW
v BW  4 m / s, vWE  3 m / s
VBE
v BE  v BW  v WE (vector addition )
Since the boat and water velocities are
perpendicular,
v BE  4 2  32  5 m / s
If the river is 50 m across, how far downstream does the boat land on the other side of the
river?
t
-
y
v BW

50
 12.5 s, x  vWE t  (3)(12.5)  37.5 m
4
In what direction should the nose of the
boat be pointed so that it goes straight
across the river?
v 
3
  tan  WE   tan    36.9 o
4
 v BE 
VWE
VBE
VBW
(upstream)
-
What is the speed of the boat with
respect to the water?
v BE  v BW 2  vWE 2  4 2  32  2.65 m / s
-
How long does it take the boat to cross the river?
t
y
v BE

50
 18.9 s
2.65
7