fisica1-youn-e-freedman-exercicios-resolvidos

1.70:
The third leg must have taken the sailor east a distance
and a distance north
o
( 5 .80 km) − ( 3.50 km) cos 45 − ( 2.00 km) = 1.33 km
o
( 3 .5 km) sin 45 = ( 2.47 km)
The magnitude of the displacement is
(1.33 km)
2
+ (2.47 km)
2
= 2.81km
.47
and the direction is arctan ( )
2
1.33
= 62 ° north of east, which is 90° − 62°
= 28°
east
of north. A more precise answer will require retaining extra significant figures in
the intermediate calculations.
1.71: a)
b) The net east displacement is
o
o
o
− 2 .80 km sin 45 + 7.40 km cos 30 − 3.30 km cos 22 = 1.37
( ) ( ) ( ) km, and the net north
o
o
o
displacement is − ( 2 .80 km) cos 45 + ( 7.40 km) sin 30 − ( 3.30 km) sin 22.0 = 0.48 km,
and so the distance traveled is ( 1.37 km) + ( 0.48 km) = 1.45 km.
2
2