CH 4 – Motion in Two and Three Dimensions
Displacement, velocity, and acceleration
Displacement: r  r f  ri
Average velocity:
v
r
t
 r dr

dt
 t 0  t
y
Instantaneous velocity: v  lim
r
Velocity is always in a direction tangent to the path
v
a
t
Average acceleration:
Instantaneous acceleration:
 v dv

dt
 t 0  t
a  lim
If a is constant (in magnitude and direction), then
v = v0 + at
r = r0 + v0 t + 1 at 2
2
For 2D motion,
v y = v0 y + a y t
x = x0 +v0xt + 1 axt 2
2
2
vx2 = v0x
+ 2ax (x - x0 )
where
y = y0 + v0 y t + 1 a y t 2
2
2
2
v y = v0y + 2a y (y - y0 )
vx0  v0 cos 0 and v y0  v0 sin0
Projectile motion
For projectile motion,
rf
x
Constant Acceleration
vx = v0x + axt
ri
a x  0 and a y   g.
So,
1
x=x0 +vx0t
x-direction:
vx0 =v0 cosθ0
vx =vx0 = const
v y0 =v0 sinθ0
y=y0 +v y0 t- 1 gt 2
2
y-direction: v y =v y0 -gt
v y 2 =v y0 2 -2g(y-y0 )
total velocity:
-
v= vx 2 +v y 2
θ=tan-1(v y /vx )
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 
-
g
(finish)
Range of a projectile over flat ground.
Δx=vx0 t=
-
2v y0
2vx0 v y0
g
2v0 2 sinθ0 cosθ0
=
g
Maximum height of a projectile. At peak height
v y  v y0
 v y0  1  v y0 
  2 g

 gt  0, t 
, y  v y 0 t  gt  v y 0 

 g 
g
 g 


Or, Δy=
v y0
2
2
1
2
v y0 2
2g
Example: A football is kicked from the ground with an initial speed of 25 m/s at an angle of 30o
above the ground.
vx0 =v0 cosθ0 =(25 m/s)cos(30 o )=21.7 m/s
v y0 =v0 sinθ0 =(25 m/s) sin(30 o )=12.5 m/s
2v y0
2(21.7)
=4.43 s
g
9.8
range =vx0 t=(12.5)(4.43)=55.4 m
hang time =
=
v y0 2
(12.5)2
maximum elevation=
=
=8.0 m
2g
2(9.8)
What is the speed and direction of motion of the ball 2 s after it is kicked?
2
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)
2
 b  b 2  4ac 12.5  12.5  4(4.9)( 20) 12.5  23.4
t


 3.66 s,  1.10 s
2a
2(4.9)
9.8
-
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?
3
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
ac 
v2
.
r
v1
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
(for small )
s v

r
v
vs
v 
r
v2
s

v2
r
v = |v1| = |v2|
 
ac 
v1
v 
v
ac
ac
v
2
v v s v
v

 v
t r t r
r
ac
v
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
4
Combined tangential and centripetal acceleration
at
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
dv
dt
ac
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
a  at 2  ac 2 , where at 
dv
v2
and ac 
dt
r
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
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
v BE  v BW  vWE (vector addition )
Since the boat and water velocities are
perpendicular,
v BE  4 2  32  5 m / s
5
VBE
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?
VWE
VBE
 vWE
 v BE

3
  tan    36.9 o
4

(upstream)
VBW
  tan 
-
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
6