boat time
Suppose you are anchored near the shore of a channel in which there is steady current, and
you are going to run your (motor)boat at constant throttle to a dock directly across the
channel on the opposite shore. There are two ways one might steer the boat to the dock:
•
the crabbing method: steer a steady course with the nose of the boat
pointed somewhat upstream, so the boat maintains a fixed orientation
and crabs in a straight line across the channel
•
the pointing method: keep the nose of the boat pointed directly at the
dock
dock
dock
start
crabbing method
start
pointing method
current
Which method gets the boat to the dock faster, and by how much? (Assume the boat
runs at a constant speed relative to the water, which is faster than the speed of the
current relative to the shore.)
Diego Scarabelli’s Solution
The Crabbing Method
For the crabbing method we have the following situation,
finish
L
vr
vb
Thus, vb = vr2 − vw2 , so the time it takes the boat to
vw
start
where vb is the velocity of the boat, vw is the velocity of the
water, vr is the velocity of the boat relative to the water,
and L is the distance between the shores.
cross the channel using the crabbing method is
.
Tc =
L
L
=
.
2
vb
vr − vw2
Copyright © 2000-2005 Michael A. Gottlieb. All rights reserved.
(1)
The Pointing Method
For the pointing method it is more convenient to consider the water to be still, and the finish point
to be moving at constant speed in the direction opposite the current. We use a coordinate system
with the start point as origin (note: positive x is leftward):
y
(D, L)
Here, vw is the velocity of the finish
point, vr is the velocity of the boat, θ
is the angle between vr and the yaxis, and (D, L) are the coordinates
of the finish point and the boat
when they meet, after an interval of
time Tp .
vw
finish
vr cos θ
vr
L
θ
vr sin θ
(0,0)
x
start
The x-displacement of the boat during an interval dt equals dx = vr sin θ dt , so
(2)
T
D = ∫ vr sin θ dt
0
And the finish point moves at constant speed, so
(3)
T
D = ∫ vw dt = vw Tp
0
vw dt
The decrease in the distance between the boat and the
finish point during the interval dt is vr dt − vw sin θ dt .
Since this distance is initially equal to L,
θ
–vw sin θ dt
T
L = ∫ vr − vwsin θ dt.
0
vr d t
θ
Setting k = vr vw , we can rewrite this integral as
vr
sin θ dt
k
T
1 T
= k ∫ vw dt − ∫ vr sin θ dt.
0
k 0
T
L = ∫ kvw −
0
(4)
Substituting eqs. (2) and (3) into (4),
L = kD −
so Tp =
Lvr
.
v − vw2
v v 
 v 2 − vw2 
D
1

= vwTp  k −  = vwTp  r − w  = Tp  r
,
k
k
v
v
v

r 

r

 w
(5)
2
r
From eqs. (1) and (5), Tc Tp = 1 − ( vw vr ) < 1 . We therefore conclude that the crabbing
2
method is faster than the pointing method.
Copyright © 2000-2005 Michael A. Gottlieb. All rights reserved.