P1112
Fall 2008
Homework #2
Solutions
1.
For the car to stop
0
)
(
=
stop
t
v
. Thus using the equation for constant
acceleration
( )
at
v
t
v
+
=
0
we get
a
v
t
t
a
v
stop
stop
0
0
0
=
!
"
=
.
Now if
2
/
a
a
!
and
0
0
2
v
v
!
then
stop
stop
t
a
v
t
4
2
2
0
=
!
2.
Data
( )
( )
j
i
t
v
j
i
t
v
s
m
s
m
s
m
s
m
ˆ
40
ˆ
170
,
ˆ
110
ˆ
90
2
1
+
!
=
+
=
r
r
(a) The two vectors differ by
( )
( )
( )
( )
(
)
( )
( )
(
)
j
i
j
i
v
j
t
v
t
v
i
t
v
t
v
t
v
t
v
v
s
m
s
m
s
m
s
m
s
m
s
m
y
y
x
x
ˆ
70
ˆ
260
ˆ
110
40
ˆ
90
170
ˆ
ˆ
1
2
1
2
1
2
!
!
=
"
#
$
%
&
’
!
+
"
#
$
%
&
’
!
"
#
$
%
&
’
!
=
(
)
!
+
!
=
!
=
(
r
r
(b) By definition
t
v
a
avg
!
!
=
r
r
and
s
t
30
=
!
so
r
a
avg
=
"
260
m
ˆ
i
"
70
m
ˆ
j
30
s
=
"
8.67
ˆ
i
"
2.33
ˆ
j
(c) The magnitude
a
avg
=
x
,
avg
2
+
a
y
,
avg
2
=
8.67
m
2
+
2.33
m
2
=
9
m
The angle is given by
( )
!
"
#
$
%
&
==
’
=
x
y
x
y
a
a
a
a
arctan
tan
(
As will be the
case here it is important to remember that arc tangent is well define up
to a factor of
o
180
.
arctan
2.33
8.67
"
#
$
%
&
’
=
15
o
of course both the x and y
components are negative so the angle is really
"
=
15
o
+
180
o
=
195
o
3.
Going back to a formula for constant acceleration
( )
( )
2
0
v
t
v
t
v
avg
+
=
. The Car
is moving at constant velocity so
t
v
x
C
C
!
=
!
. In order for the motorcycle to
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 Fall '07
 LECLAIR,A
 Acceleration, Work, 9m, #XM, #xC

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