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2
2
7.00 18.0 ,
18.0
dy
d y
t
dt
dt
=−
=
−
(a) The acceleration is
22
ˆˆ
ˆ
ˆ
ˆ
ˆ
i
j
i
j
( 24.0 )i ( 18.0)j.
xy
dx
dy
aa a
t
dt
dt
=+=
+
=
−
+
−
G
At
0.700 s
t
=
, we have
( 16.8)i ( 18.0)j
a
+−
G
with a magnitude of
2
 
( 16.8)
( 18.0)
24.6 m/s .
aa
==
−
+
−
=
G
Thus, the magnitude of the force is
2
(0.34 kg)(24.6 m/s )
8.37 N.
Fm
a
=
(b) The angle
F
G
or
/
aFm
=
G
G
makes with
x
+
is
2
11
2
18.0 m/s
tan
tan
47.0 or 133 .
16.8 m/s
y
x
a
a
θ
−−
§·
−
=
°
−
°
¨¸
−
©¹
We choose the latter (
133
−°
) since
F
G
is in the third quadrant.
(c) The direction of travel is the direction of a tangent to the path, which is the direction
of the velocity vector:
2
ˆ
ˆ
ˆ
ˆ
( )
i
j
i
j
(2.00 12.0 )i (7.00 18.0 )j.
dx
dy
vt
v
v
t
t
dt
dt
+
=
−
+
−
G
At
0.700 s
t
=
, we have
(
0.700 s)
( 3.88 m/s)i ( 5.60 m/s)j.
−
+
−
G
Therefore, the angle
v
G
makes with
x
+
is
5.60 m/s
tan
tan
55.3 or 125 .
3.88 m/s
y
v
x
v
v
−
=
°
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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, Force

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