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3.45:
a) The
0
=
x
a
and
β
a
y
2

=
, so the velocity and the acceleration will be
perpendicular only when
0
=
y
v
, which occurs at
0
=
t
.
b) The speed is
0
/
,
)
4
(
2
/
1
2
2
2
=
+
α
=
dt
dv
t
β
v
at
0
=
t
. (See part d below.)
c)
r
and
v
are perpendicular when their dot product is 0:
0
2
m)
0
.
30
(
)
2
(
)
m
0
.
15
(
)
)(
(
3
2
2
2
=
+

α
=

×

+
α
α
t
β
βt
t
βt
βt
t
. Solve this for
t
:
s
208
.
5
2
2
2
2
)
m/s
500
.
0
(
2
m/s)
2
.
1
(
)
m/s
m)(0.500
0
.
30
(
+
=
±
=

t
, and 0 s, at which times the student is at (6.25 m,
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Unformatted text preview: 1.44 m) and (0 m, 15.0 m), respectively. d) At s 208 . 5 = t , the student is 6.41 m from the origin, at an angle of ° 13 from the xaxis. A plot of 2 / 1 2 2 ) ) ( ) ( ( ) ( t y t x t d + = shows the minimum distance of 6.41 m at 5.208 s: e) In the x y plane the student’s path is:...
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