• Ch 1 Mathematical Concepts
• Dimensional Analysis
Review for Exam 1
• Trigonometry
• Vectors, Addition and Subtraction, Components
• Addition of Vectors by Means of Components
• Ch 2 Kinematics in One Dimension
Practice Problem Solving !
Homework Problems
Recitation Problems
Examples from Class
Adding Vectors
• Displacement, Speed, Velocity, Acceleration
• Equations of Kinematics for Constant Acceleration
• Applications; Freely Falling Bodies
• Ch 3 Kinematics in Two Dimensions
• Displacement, Velocity, Acceleration (Vectors !)
• Equations of Kinematics in Two Dimensions
• Projectile Motion
• Relative Velocity
Example: Displacement of a cross-country skier
On a cross-country ski trip
you travel 1.00 km north and
then 2.00 km east.
a) How far and in what
direction are you from your
starting point ?
Components of Vectors
d = (1.00km) 2 + (2.00km) 2
= 2.24 km
(magnitude of the displacement vector)
tan φ =
opposite side 2.00 km
=
adjacent side 1.00 km
φ = 63.4o
(direction of the displacement vector)
Equations of Kinematics in One Dimension
For constant acceleration
we find:
Position
x
Velocity
r
r
Δx
v = lim
Δt →0 Δt
t
v
1
x = x0 + v0t + at 2
2
t
a
a = const
v2 − v0 = 2a (x − x0 )
2
s2
Find the position of the accelerating speedboat at t = 8.0 s.
x = v o t + 12 at 2
t
g = 9 .8 m
vo = 6.0 m/s
a = +2.0 m/s2
v = v 0 + at
Acceleration
r
r
Δv
a = lim
Δt →0 Δt
Equations of kinematics for constant acceleration
(
)
= (6.0 m s ) (8.0 s ) + 12 2.0 m s 2 (8.0 s )
2
= +110 m
1
Velocity in two dimensions
Definition of
average velocity
v av =
Acceleration in two dimensions
r2 − r1 Δ r
=
t 2 − t 1 Δt
Definition of average acceleration
a av =
Definition of
instantaneous velocity
v = lim
Δr
Δt
v 2 − v1 Δ v
=
t 2 − t1
Δt
Definition of instantaneous acceleration
a = lim
Most important concept in two-dimensional motion
• two-dimensional motion can be decomposed into motion
in x-direction and motion in y-direction
“x” and “y” components of motion are independent
Constant-Acceleration
Equations of Motion in
Two-Dimensions
vx = v0x + axt
x = x0 + v0xt + (½ )ax t2
• Person on a cart throws ball vertically upwards
– Seen from two different frames of reference
Reference frame
on moving cart
Equations for Projectile
Motion (assuming that
ax = 0, ay = -g)
vx = v0x
x = x0 + v0xt
(1)
(2)
vy = v0y + ayt
y = y0 + v0yt + (½ )ay t2
vy = v0y – gt
(3)
2
y = y0 + v0yt - (½ )gt (4)
v2x − v0x = 2a x (x − x0 )
v2y −v0y = −2g (y− y0)
2
Reference frame
on the ground
Δv
Δt
v2y − v0y = 2ay (y − y0 )
2
2
Example: A moving spacecraft
Example: The height of a kickoff
A placekicker kicks a football at and angle of 40.0 degrees and the
initial speed of the ball is 22 m/s. Ignoring air resistance,
determine the maximum height that the ball attains.
v=
v
θ
v x = 190 m s
(190
m s ) + (98 m s )
2
2
= 210 m s
v y = 98 m s
θ = tan −1 (98 190 ) = 27 o
vo
θ
v ox = v o cos θ = (22 m s ) cos 40 o = 17 m s
v oy
Voy = Vo sin θ = (22m / s )(sin 40°) = 14 m / s
v ox
2
Relative velocity in 2-dim
The concept of relative
velocity can be extended
from 1-dim to 2-dim
y
?
ay
-9.80
m/s2
vy
voy
0
14 m/s
2
v 2y = v oy
+ 2a y y
0 − (14 m s )
y=
= +10 m
2 − 9 .8 m s 2
2
(
)
y=
t
2
v 2y − v oy
2a y
Velocities can carry multiple values depending on the position
and motion of the object and observer
3