Physics 1
Review
Name _________________________
Period _____
1.
Uncertainty in measurement
8.

Data "correctness" is measure by % difference:
% = 100|mean – true|/true
Data "precision" is measured by % deviation:
%  = 100 |trial – mean|/N(mean)
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
2.
Fundamental units in physics


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3.
Measure mass in kilograms
Measure length in meters
Measure time in seconds
Displacement

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
4.
Difference in position over a period of time t : d = x – xo
Is measured in meters (m)
Distance is measured along path taken and is positive
Graphing displacement for one dimensional motion

Direction graphed along x-axis
backward (–)
(+) forward
0
Position vs. time graph: Position along the y-axis and time
along the x-axis (time is always positive)
+x
moving in the forward
direction has a positive
slope

0
moving in the backward
direction has a negative
slope
5.
Average velocity



6.
Displacement divided by time: vav = (x – xo)/t = d/t
Is measured in meters/seconds (m/s)
Speed is distance divided by time and is always positive
Acceleration


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Change in velocity divided by time: a = (v – vo)/t
Is measured in (meters/seconds)/seconds (m/s2)
Course limited to zero acceleration (constant motion) or
constant acceleration (accelerated motion)
When an object accelerates from rest, then final velocity is
twice average velocity: v = 2vav = 2(d/t)
7.
Acceleration due to gravity: An object thrown up into the air
and falls back down

Acceleration due to gravity (g) is constant and downward
(whether moving up, at the top of the arc, or moving down)
g = -9.8 m/s2 or  -10 m/s2
Graphing position vs. time
Graphing velocity vs. time
d
v
-g
+vo leave your hand


-g
-g
0
-vo
-g
Positive displacement and velocity indicate that the object
is moving in the defined forward direction (negative is in
the opposite direction)

displacement and velocity always have the same sign

When acceleration has the same sign as velocity, the
object is speeding up (opposite signs = slowing down)
9. Kinematic Equations
Formula (vo = 0)
Unused Variable
Formula (vo  0)
v
d = ½at2
d = vot + ½at2
a
d = ½vt
d = ½(vo + v)t
d
v = at
v = vo + at
t
v2 = 2ad
v2 = vo2 + 2ad
10. Derive v2 = vo2 + 2ad
solve for t:
t = 2d/(vo + v)
t = (v – vo)/a
combine:
2d/(vo + v) = (v – vo)/a
cross multiply:
2ad = (v + vo)(v – vo)
foil:
2ad = v2 – vo2
11. Graphing one dimensional motion.

constant velocity graphs
d
v
a
t
-x

Positive and Negative Case
½t
-g
t
highest point
½t
t
return to hand
t

d
t
constant acceleration graphs (a  0)
v
t
t
a
t
12. solving kinematics problems

draw diagram

constant motion (a = 0)
complete chart with two numbers and one letter
d
vav
t

accelerated motion (a  0)
complete chart with given information
(minimum requirement: three numbers and two letters)
d
vo
v
a
t
13. Displacement equals area under velocity vs. time graph
and acceleration equals slope of velocity vs. time graph
What is the distance traveled
v
in the first four seconds?
d = vavt
d = ½(20 m/s)(4 s) = 40 m
What is the acceleration during
the first four seconds?
a = v = 20 m/s – 0 m/s
t
4s
a = 5 m/s2
t
t