2 A watermelon seed has the following coordinates: x = -5.0 m, y = 9.0 m, and = = 0 m. Find its position vector (a) in unit-vector no- tation and as (b) a magnitude and (c) an angle relative to the positive direction of the r axis. (d) Sketch the vector on a right-handed coor- dinate system. If the seed is moved to the xyz coordinates (3.00 m, 0 m.O m), what is its displacement (e) in unit-vector notation and as f) a magnitude and (g) an angle relative to the positive x direction?
Answers
Answer:
a) The position vector, according to Eq. 4-1, is r = (−5.0m)
i
^
+(8.0m)
j
^
(b) The magnitude is ∣r∣ =
x
2
+y
2
+z
2
=
(−5.0m)
2
+(8.0m)
2
+(0m)
2
= 9.4 m.
(c) Many calculators have polar ↔ rectangular conversion capabilities that make this computation more efficient than what is shown below. Noting that the vector lies in the xy plane and using Eq. 3-6, we obtain:
θ=tan
−1
(
−5.0m
8.0m
)=−58
o
or122
o
where the latter possibility (122° measured counterclockwise from the +x direction) is chosen since the signs of the components imply the vector is in the second quadrant.
(d) The sketch is shown to the right. The vector is 122° counterclockwise from the +x direction.
(e) The displacement is Δ
r
=
r
−
r
where
r
is given in part (a) and
r
=(3.0m)
i
^
. Therefore, Δ
r
=(8.0m)
i
^
−(8.0m)
j
^
.
(f) The magnitude of the displacement is ∣Δ
r
∣=
(8.0m)
2
+(−8.0m)
2
=11m.
(g) The angle for the displacement, using Eq. 3.6, is tan
−1
(
−8.0m
8.0m
)=−45
o
or135
o
where we choose the former possibility (−45°,or45° measured clockwise from +x) since the signs of the components imply the vector is in the fourth quadrant. A sketch of Δ
r
is shown on the right.
