Motion in Two or Three Dimensions
331
0
(30.0 m/min)(4.00 min)
120.0 m
xx
−=
=
. You will land 120.0 m east of point
B
, which is 45.0 m east of
point
C
. The distance you will have traveled is
22
(400.0 m)
(120.0 m)
418 m
+=
.
(b)
B/W
v
!
is directed at angle
φ
east of north, where
75.0 m
tan
400.0 m
=
and
10.6
=
°
.
B/W
(100.0 m/min)sin10.6
18.4 m/min
x
v
==
°
and
B/W
(100.0 m/min)cos10.6
98.3 m/min
y
v
°
.
B/E
B/W
W/E
18.4 m/min
30.0 m/min
48.4 m/min
xxx
vvv
=+=
+
=
.
B/E
B/W
W/E
98.3 m/min
yyy
.
0
B/E
400.0 m
4.07 min
98.3 m/min
y
yy
t
v
−
=
.
0
(48.4 m/min)(4.07 min)
197 m
=
. You will land 197 m downstream
from
B
, so 122 m downstream from
C
.
(c)
(i) If you reach point
C
, then
B/E
v
!
is directed at 10.6
°
east of north, which is 79.4
°
north of east. We don’t know
the magnitude of
B/E
v
!
and the direction of
B/W
v
!
. In part (a) we found that if we aim the boat due north we will land
east of
C
, so to land at
C
we must aim the boat west of north. Let
B/W
v
!
be at an angle
of north of west. The
relative velocity addition diagram is sketched in Figure 3.76. The law of sines says
W/E
B/W
sin
sin79.4
vv
θ
=
°
.
30.0 m/min
sin
sin79.4
100.0 m/min
⎛⎞
=
⎜⎟
⎝⎠
°
and
17.15
=
°
. Then
180
79.4
17.15
83.5
=−
−
=
°°
. The boat will head
83.5
°
north of west, so 6.5
°
west of north.
B/E
B/W
W/E
(100.0 m/min)cos83.5
30.0 m/min
18.7 m/min
−
+
=
°
.
B/E
B/W
W/E
(100.0 m/min)sin83.5
99.4 m/min
−
=
°
. Note that these two components do give the
direction of
B/E
v
!
to be 79.4
°
north of east, as required. (ii) The time to cross the river is
0
B/E
400.0 m
4.02 min
99.4 m/min
y
t
v
−
=
. (iii) You travel from
A
to
C
, a distance of
(400.0 m)
(75.0 m)
407 m
.
(iv)
B/E
B/E
B/E
(
)
(
)
101 m/min
xy
v
=+
=
. Note that
B/E
406 m
vt
=
, the distance traveled (apart from a small
difference due to rounding).
EVALUATE:
You cross the river in the shortest time when you head toward point
B
, as in part (a), even though
you travel farther than in part (c).
Figure 3.76
3.77.
IDENTIFY:
/
x
vd
x
d
t
=
,
/
y
y
d
t
=
,
/
ad
t
=
and
/
t
=
.
SET UP:
(sin
)
cos(
)
dt
t
dt
ω
=
and
(cos
)
sin(
)
t
dt
.
EXECUTE:
(a)
The path is sketched in Figure 3.77.
(b)
To find the velocity components, take the derivative of
x
and
y
with respect to time:
(1 cos
),
x
vR
ω
t
and
sin
.
y
ω
t
=
To find the acceleration components, take the derivative of
x
v
and
y
v
with respect to time:
2
sin
x
aR
t
,
=
and
2
cos
y
t
.
=
(c)
The particle is at rest (
0)
yx
every period, namely at
0 2 /
4 /
....
t,
πω
,
,
=
At that time,
0 2
4
.
.
.
;
x,
π
R,
π
R,
=
and
0.
y
=
The acceleration is
2
=
in the

y
+
direction.