fisica1-youn-e-freedman-exercicios-resolvidos

1.78: (a) Take the beginning of the journey as the origin, with north being the y-
direction, east the x-direction, and the z-axis vertical. The first displacement is then
− 30kˆ,
the second is − 15 ĵ , the third is 200 i ˆ (0.2 km = 200 m) , and the fourth is 100 ĵ .
Adding the four:
− 30kˆ
−15 ˆj
+ 200ˆ i + 100 ˆj
= 200ˆ i + 85 ˆj
− 30kˆ
(b) The total distance traveled is the sum of the distances of the individual segments: 30
+ 15 + 200 + 100 = 345 m. The magnitude of the total displacement is:
D
D
D
D
2 2 2
=
x
+
y
+
z
=
200
2
+ 85
2
+
2
( − 30) = 219m
1.79: Let the displacement from your camp to the store be A r
.
A =
r
B is 32
240 m, 32
o
south of east
r
south of west and C
Let + x be east and + y be north
r r r
A + B + C = 0
A
A
x
y
+ B
+ B
o
x
y
+ C
x
+ C
y
o
= 0, so A cos 32
is 62
o
= 0, so − A sin 32
o
south of west
− B cos 48
o
+ B sin 48
− C cos 62
o
o
− C sin 62
A is known so we have two equations in the two unknowns B and C. Solving gives
B = 255 m and C = 70 m.
= 0
o
= 0
1.80: Take your tent's position as the origin. The displacement vector for Joe's tent is
o
( ) ˆ
o
21cos
23 i − ( 21sin 23 ) ˆj
= 19.33ˆ i − 8.205 ˆj
. The displacement vector for Karl's tent is
o
( ) iˆ
o
32 cos 37 + ( 32 sin 37 ) ˆj
= 25.56ˆ i + 19.26 ˆj
. The difference between the two
displacements is:
( 19.33
25.56) iˆ
+ ( − 8.205 −19.25) ˆj
= −6.23ˆ
i − 27.46 ˆj
− .
The magnitude of this vector is the distance between the two tents:
D =
2
2
( − 6.23) + ( − 27.46) = 28.2 m