fisica1-youn-e-freedman-exercicios-resolvidos

1.39: Using components as a check for any graphical method, the components of B r are
B = 14.4 m and B = 10.8 m, A r has one component, A = −12 m .
x
y
a) The x - and y - components of the sum are 2.4 m and 10.8 m, for a magnitude
2
2
⎛
+ , and an angle of 10.8 ⎞
⎜ ⎟ = 77.6
o .
⎝ 2.4 ⎠
b) The magnitude and direction of A + B are the same as B + A.
c) The x- and y-components of the vector difference are – 26.4 m and
− 10.8
o
−10.8 m, for a magnitude of 28.5 m and a direction arctan ( ) = 202 . Note that
of ( 2.4 m) ( 10.8 m) = 11.1m,
o
10 .8
10.8 o
180 must be added to arctan( ) = arctan( ) = 22
−26.4
26. 4
third quadrant.
Magnitude =
x
−26.4
−
in order to give an angle in the
r
2 2
⎛ 10.8 ⎞ o
( 26.4 m) ( 10.8 m) 28.5 m at and angle of arctan = 22.2 .
d) B − A
r
= 14.4 mˆ i + 10.8 mˆj
+ 12.0 mˆ i = 26.4 mˆ i + 10.8 mˆj
.
+
=
⎜ ⎟
⎝ 26.4 ⎠
1.40: Using Equations (1.8) and (1.9), the magnitude and direction of each of the given
vectors is:
a)
180 o – 31.2 o ).
b)
c)
360 o – 19.2 o ).
( +
2
2
5.20
− 8.6 cm) (5.20 cm) = 10.0 cm, arctan ( )
8.60
( −
2
2
2.45
− 9.7 m) + ( 2.45 m) = 10.0 m, arctan ( )
−9.7
( 7.75 km) −
2
2
2.7
+ ( 2.70 km) = 8.21 km, arctan ( )
7.75
−
= 148.8 o (which is
−
= 14 o + 180 o = 194 o .
−
= 340.8 o (which is