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()
2
ˆˆ
ˆ
6.0
4.0
i + 8.0 j
6.0
8.0 i
dv
d
at
t
t
dt
dt
==
−
=
−
G
G
in SI units. Specifically, we find the acceleration vector at
3.0 s
t
=
to
be
2
6.0 8.0(3.0) i ( 18 m/s )i.
−=
−
(b) The equation is
G
=−
60 80
..
#
bg
i=0
; we find
t
= 0.75 s.
(c) Since the
y
component of the velocity,
v
y
= 8.0 m/s, is never zero, the velocity cannot
vanish.
(d) Since speed is the magnitude of the velocity, we have

vv
=
G
2
2
2
6.0
4.0
8.0
10
tt
+=
in SI units (m/s). To solve for
t
, we first square both sides of the above equation, followed
by some rearrangement:
22
6.0
4.0
64
100
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Acceleration

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