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PHSX 114 – Spring 2008
Formulae for 4
th
Exam
Constant
: g = 9.80 m/s
2
downwards
Trigonometric
Relations
:
Kinematic Formulae
:
t
x
t

t
x

x
v
1
2
1
2
ave
∆
∆
=
=
t
v
t

t
v

v
a
1
2
1
2
ave
∆
∆
=
=
t
x
0
t
lim
v
∆
∆
→
∆
=
t
v
0
t
lim
a
∆
∆
→
∆
=
at
v
v
0
+
=
2
0
0
at
2
1
t
v
x
x
+
+
=
)
x
2a(x
v
v
0
2
0
2

+
=
2
v
v
v
0
ave
+
=
Distance traveled,
t
v
x

x
ave
0
=
For motion in earth’s gravitational field with the +y axis chosen to be upwards,
gt
v
v
0

=
2
0
0
gt
2
1
t
v
y
y

+
=
)
y
2g(y
v
v
0
2
0
2
2


=
,
2
v
v
v
0
ave
+
=
but you must be careful when you use the latter relation
Vectors
:
With A = magnitude of
A
and θ = angle between the
direction of
A
and the xaxis,
A
x
= Acos θ,
A
y
= A sin θ,
A
2
= A
x
2
+ A
y
2
,
sin θ = cos Φ , cos θ = sin Φ
θ = tan
1
x
y
A
A
For
B
A
C
+
=
, etc.,
C
x
= A
x
+ B
x
and
C
y
= A
y
+B
y
Quadratic equation
:
Solution to the quadratic equation, ax
2
+ bx + c = 0 is:
2a
4ac
b
b

x
2

±
=
Phsx114sp08–Formula Sheet for Exam 4, Page 1 of 3
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This note was uploaded on 04/09/2008 for the course PHSX 114 taught by Professor Davis during the Spring '08 term at Kansas.
 Spring '08
 DAVIS

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