AP Physics Exam 1 Review
Chapter 1 - 4
1
1mile  1600m
1hr  3600s
Conversion Factors
2 mile/hr = __ m/s
mile
mile  1600m   1hr   0.89 m
2
2
 
 


s
3600
s

hr
hr
 mile  
2
Significant Figures (Digits)
1. Nonzero digits are always significant.
2. The final zero is significant when there is a decimal
point.
3. Zeros between two other significant digits are always
significant.
4. Zeros used solely for spacing the decimal point are not
significant.
Example:
 1.002300  7 sig. fig’s

3
0.004005600  7 sig. fig’s

12300  3 sig. fig’s
Addition and subtraction with Sig. Figs

The sum or difference of two measurements is
precise to the same number of digits after the
decimal point as the one with the least number
of digits after the decimal point.
Example:
16.26 + 4.2 = 20.46 =20.5
4
Multiplication and Division with Sig.
Figs

The number of significant digits in a product or quotient
is the number in the measurement with the least
number of significant digits
Example:
2.33  5.5 = 12.815 =13.
5
Position and Displacement
x = x2 - x1 = xf – xi
Vector
–
–
6
Magnitude: how far
Direction: Negative sign indicates direction only, it has nothing
to do with magnitude.
Total Distance, dtot

When there is no change in direction:
d  x

When there is change in direction:
dtotal  d1  d 2  ...  x1  x2  ...
where d1 and d2 are distances of segments in which
there is no change in direction.
7
Average Velocity and Speed
x x f  xi
v

t
t f  ti
dtotal
s
t
8
 x  v  t

Standard Unit: m/s

Constant velocity: v  v
Instantaneous velocity

Instantaneous velocity is the average velocity
when the time interval becomes very, very small,
essentially zero.
x dx
v  lim

t  0 t
dt

9
Instantaneous velocity is the time-derivative of
position function.
Acceleration

Average acceleration
v v f  vi
a

t
t f  ti

10
Instantaneous acceleration
v dv
a  lim

t  0 t
dt
Constant Acceleration Motion


11
a = constant
Let initial time
t = 0, then at
any time t,
velocity and
position are
given by

v  v0  at

 x  x  v t  1 at 2
0
0

2
 2
2
v

v

0  2a  x  x0 

 x  x0  1  v0  v   t  v  v0  v
2
2


1 2
 x  x0  vt  at
2

Free-Fall Motion
a = g, downward (g = 9.81 m/s2)

Up, down or on top of path
Acceleration can be + or g always +

g is acceleration due to gravity (It is not gravity.)


12
Adding Vectors: Head-to-Tail

Head-to-Tail method:
Example: A + B
–
–
–
A
B
Draw vector A
Draw vector B starting
from the head of A
The vector drawn from the
tail of A to the head of B is
the sum of A + B.
B
A
13
Vector Components

 ax  a cos 

a y  a sin 



 is the angle between the
vector and the +x axis.

ax and ay are scalars.
y
a
ay

ax
14
x
y
ay
Vector magnitude
and direction
 magnitude: a  a 2  a 2
x
y



ay
a
1 y
 tan  
 direction:   tan
a

ax
x

 is the angle from the +x axis to the vector.
15
a

ax
x
y
ry
Adding Vectors
by Components

ay
b
by
a
r
ax
rx
When adding vectors by components, we add
components in a direction separately from other
components.
r  a  b
x
x
x
r  ab


  ry

rz


2-D
 a y  by
 az  bz
3-D
Component form:
r  a  b   ax  bx  iˆ  ay  by ˆj   az  bz  kˆ
16
bx

s  a  b   a  b  iˆ   a
x
x
y

 b  ˆj   a
y
z
 bz  kˆ
x
Dot product: ab
b
a
a  b  ab cos 

ba is the projection of b onto a.
ba  b cos 
 a  b  ab cos   aba
b

ba
a b  axbx  a yby  az bz
17
a
A B
Cross Product: c = a  b
A
c  ab sin 

Magnitude of c is:

c is a vector, and it has a direction given by the
right-hand-rule (RHR):
–
–
–
–
Place the vectors a and b so that their tails are at the same point.
Extend your right arm and fingers in the direction of a.
Rotate your hands along your arm so that you can flap your
fingers toward b through the smaller angle between a and b.
Then
Your outstretched thumb points in the direction of c.
 
a  b  a y bz  az by iˆ  az bx  axbz  ˆj  axby  a y bx kˆ
18
Position is a 3D vector
k
 r  x2  y2  z 2


1 y
  tan x

2
2

x

y
  tan 1
z
z

y
x
i
19
r

x2  y2
j
3-D Velocity and Acceleration
The instantaneous
velocity v of a particle is
always tangent to the
path of the particle.
20

dx
 vx 
dt

dy

 vy 
dt

dz

 vz  dt


dvx
 ax 
dt

dv y

  ay 
dt

dvz

a

 z
dt

Projectile Motion Breakdown

Horizontal: constant velocity
 vx  vox
 x x v t
o
ox


Vertical: Constant acceleration (ay = g, downward)
 ay = g if downward is defined as +y direction
 ay = -g if upward is defined as +y direction
 v y  voy  a y t
 y  y  v t  1 a t2
o
oy
y

2
 vy 2  voy 2  2ay y

21

 v0 x  v0 cos 0

v0 y  v0 sin 0


Height, Range and Equation of Path

22
Minimum speed at top, but  0
 vy = 0
 vx = v0x = v0 cos 0  vmin  vo cos  o

Maximum Height is
v02 sin 2  0
H
2g

Horizontal range is
v0 2
R
sin  20 
g

Equation of path is
y   tan 0  x 
gx 2
2  v0 cos 0 
2
Centripetal Acceleration
v2
a
r

av
for uniform circular
motion at any time.
v
a
a
a
v
v
23
Uniform Circular Motion
Period:
2 r
T
v
Frequency:
1
f 
T
24
Relative Motion
v AC  v A B  v B C
v AB  vBA
25