Ch 4
2-D motion
Questions: 6, 7
Problems: 1, 3, 5, 7, 11, 17, 21, 23,
25, 29, 31, 33, 39, 45, 47, 57, 61, 67
Expand our knowledge (and the rules) of 1-D
motion to two and three dimensions.
examples: throwing a ball, shooting a cannon,
a plane in flight
Position of a particle is described by the
position vector:
y

r
xiˆ
yˆj zkˆ
2 mˆj
5mkˆ
r

r
( 3m)iˆ (2m) ˆj 5mkˆ
z
3miˆ
x
Displacement – change in position

r

r

r

r
 
r2 r1
( x2iˆ
( x2
xiˆ
y2 ˆj z 2 kˆ) ( x1iˆ y1 ˆj z1kˆ)
x )iˆ ( y y ) ˆj ( z z )kˆ
1
2
yˆj
2
2
1
zkˆ
Example: Find the displacement vector

r1 (3m)iˆ (4m) ˆj (2m)kˆ

r2 (6m)iˆ (1m) ˆj (2m)kˆ
Average and instantaneous velocity.
average velocity = displacement/time

vave

vave

r
t
xiˆ
yˆj
t
zkˆ
x
i
t
y
j
t
z
k
t

Find the average velocity if r1 (3m)iˆ (4m) ˆj (2m)kˆ

r2 (6m)iˆ (1m) ˆj (2m)kˆ
and the time interval is 3 seconds.
Instantaneous velocity
The instantaneous velocity was defined as the limit of the velocity as
T approaches zero.
v
lim
t
0

r
t

dr
dt
To find the velocity vector function, take the derivative of the position
vector function as a function of time.

v

v
d ˆ ˆ
( xi yj zkˆ)
dt
v x iˆ v y ˆj v z kˆ
where: vx
dx
dt
vy
dx ˆ dy ˆ
i
j
dt
dt
dy
dt
vz
dz ˆ
k
dt
dz
dt
The direction of the instantaneous velocity vector is ALWAYS
tangent to the particle’s path at the particle’s position.
x(t)m = 2t2 -3t-2
y(t)m = -3t+2
Graph the position vector function.
Find the velocity when t = 3s.
Describe the velocity in both unit vector notation and magnitude
and direction.
The position vector graph is a parametric function.
Checkpoint 1
A particle is following a circular path. If the

instantaneous velocity is : v (2 m s )iˆ (2 m s ) ˆj
what quadrant is the particle in, if it is
traveling clockwise, counterclockwise?
y
x
Average and instantaneous acceleration
Use the same ideas to find the velocity to find
the acceleration.
 


aave
v2 v1
t
v
t
In the limit that t approaches zero, we get: a

a

a

a
d
(v x iˆ v y ˆj v z kˆ)
dt
dvx ˆ dv y ˆ dvz ˆ
i
j
k
dt
dt
dt
a x iˆ a y ˆj a z kˆ
where:

ax

dv x
dt

ay

dv y
dt

az

dv z
dt

dv
dt
The direction of the of the acceleration vector DOES NOT
necessarily point from one position to another. See figure 4.7
Checkpoint 2
We have the descriptions for position of a puck in the xy plane.
1) x=-3t2+4t-2 and y=6t2-4t

4) r (4t 3 2t )iˆ 3 ˆj
Are the x and y acceleration components constant? Is
 constant?
acceleration a
Projectile Motion
Projectile motion is probably the most used
example of motion in more than 1 dimension.
The particle will move in a vertical plane.
The particle has some initial velocity, v0.
The acceleration is the free fall acceleration, g.
We will be using 2-d motion
Assume no air resistance

v0


v 0 x i v0 y j
Where v0x = v0 cos
0
and v0y = v0 sin
0
Big rule: The horizontal and vertical motions
are independent of each other.
This is because the acceleration is from gravity
and gravity is in the vertical direction.
(Accelerations in one component do nothing to the
other component.)
See figure 4-10 to look at the velocity vectors of a
launched projectile.
Because the horizontal and vertical
components of motion are independent, you
can roll a ball off of a table at the same time
you drop another and they hit the ground at
the same time. (see fig. 4-11)
Checkpoint 3. At a certain instant, a ball has velocity

v 25iˆ 4.9 ˆj (x is horizontal, y is vertically upward).
Has the ball passed its highest point?
The horizontal velocity of a projectile under the
influence of only gravity is constant.
ax=0
x – x0 = v0xt
x – x0 = (v0cos 0)t
vx = vox = v0cos 0
0 = is the projection(initial) angle
The vertical motion is more complicated.
y – y0 = v0yt – ½ gt2
y – y0 = (v0 sin 0)t – ½ gt2
vy= v0 sin 0 – gt
vy2
(v0 sin
2
)
0
2g ( y
y0 )
These are our kinematic equations from chapter
2, written for each dimension.
Even though the horizontal and vertical
motions are independent, they do occur over
the same time.
Take the two position equations:
x – x0 = (v0cos
0)t
and y – y0 = (v0 sin
0)t
– ½ gt2
By substituting in for time, we can eliminate the time
and right an equation that describes the trajectory.
2
gx
y (tan 0 ) x
2(v0 cos 0 ) 2
Note that this is a concave down parabola.
Range equation
The horizontal distance covered is the range.
let x – x0 = R and y – y0 = 0
The second condition, means that the landing
position is at the same height as the launch.
R= (v0cos
0)t
2
)t
–
½
gt
0
and 0 = (v0 sin
eliminating t gives us:
Only use if y = y0 !!!
R
2v02
sin
g
R
v02
sin(2 0 )
g
0
cos
0
Uniform circular motion
Moving in a circle is another example of 2-D motion.
The uniform in uniform circular motion implies a constant
speed.
Even if the speed is constant, the particle is accelerating
due to the changes is direction.
The acceleration vector is pointing towards the center of
the circle. So as the particle moves around the circle, the
acceleration vector is also changing directions. Call this
acceleration: centripetal acceleration.
The centripetal acceleration is: ac
v2
r
In the time it takes for 1 revolution, the
particle will travel a distance of: 2 r
The time it takes for 1 revolution is defined as
the period.
since: v = distance/time
2 r
we can find: T
v
Checkpoint 5: An object moves at constant speed around a
circle in the xy plane centered on the origin. When the
object is at x = -2m and the velocity is ( 4m / s ) ˆj. Give the
object’s velocity and acceleration at y = 2m.
Question 4
Question 13
Problems 6, 16, 22, 26, 38, 42, 58, 60