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65. We first note that
a
1
→
(the acceleration at
t
1
= 2.00 s
) is perpendicular to
a
2
→
(the
acceleration at
t
2
=5.00 s
), by taking their scalar (dot) product.:
22
2
2
12
ˆˆ
ˆ
ˆ
[(6.00 m/s )i+(4.00 m/s )j] [(4.00 m/s )i+( 6.00 m/s )j]=0.
aa
⋅=
⋅
−
GG
Since the acceleration vectors are in the (negative) radial directions, then the two
positions (at
t
1
and
t
2
) are a quartercircle apart (or threequarters of a circle, depending
on whether one measures clockwise or counterclockwise).
A quick sketch leads to the
conclusion that if the particle is moving counterclockwise (as the problem states) then it
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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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