Exercícios halliday

8. We label the displacement vectors A
,
B and C
(and denote the result of their vector sum as r ). We choose east as the ˆi direction (+x direction) and north as the ˆj direction
(+y direction). We note that the angle between C and the x axis is 60°. Thus,
( ) ( )
(50 km) ˆi
(30 km) ˆj
(25 km) cos 60 ˆi + (25 km)sin 60 ˆj
A
B
C
=
=
= ° °

(a) The total displacement of the car from its initial position is represented by r = A + B + C =(62.5 km) ˆi + (51.7 km) ˆj which means that its magnitude is r = (62.5km)2 + (51.7 km)2 = 81 km.
(b) The angle (counterclockwise from +x axis) is tan–1 (51.7 km/62.5 km) = 40°, which is to say that it points 40° north of east. Although the resultant r is shown in our sketch, it would be a direct line from the “tail” of A to the “head” of C
.
(c) The angle between the resultant and the +x axis is given by θ = tan–1(ry/rx) = tan–1 [(10 m)/( –9.0 m)] = – 48° or 132°.
Since the x component of the resultant is negative and the y component is positive, characteristic of the second quadrant, we find the angle is 132° (measured counterclockwise from +x axis).
9. We write r = a + b . When not explicitly displayed, the units here are assumed to be meters. (a) The x and the y components of r are rx = ax + bx = (4.0 m) – (13 m) = –9.0 m and ry = ay + by = (3.0 m) + (7.0 m) = 10 m, respectively. Thus r = (−9.0m) ˆi + (10m) ˆj .
(b) The magnitude of r is
| | 2 2 ( 9.0 m)2 (10 m)2 13 m x y r = r = r + r = − + = .
10. We label the displacement vectors A
,
B and C
(and denote the result of their vector sum as r ). We choose east as the ˆi direction (+x direction) and north as the ˆj direction
(+y direction) All distances are understood to be in kilometers.
(a) The vector diagram representing the motion is shown below:
(3.1 km) ˆj
( 2.4 km) ˆi
( 5.2 km) ˆj
A
B
C
=
= −
= −

(b) The final point is represented by r = A + B + C = (−2.4 km)ˆi + (−2.1 km)ˆj