Physics - Chapter 3 
Section 3-6 - Problem 29
Problem 29:
In hot pursuit, Agent Friday* must get directly across a 1600 m wide river in the minimum amount of time.  The river current
is 0.80 m/s.  Friday can row a boat at 1.50 m/s and can run at 3.00 m/s.  Describe the path Friday should take, rowing 
plus running along the shore, and determine the minimum total time.        
We start by diagramming out the stated problem.  The diagram actually has to be drawn in two separate sections. That is the trick 
to this confusing problem. We need to point the boat (presumably UPstream) along vector VBW at the angle γ relative to  
the shore. The current pushes downstream along vector VWS. These act to create the resultant vector VBS. This is the motion
of the boat with respect to the shore.  This relationship is drawn in the right hand side of the diagram below. These are velocity
vectors. 
The angle VBS makes with a line straight across the river is identical to the angle that the boat will be traveling as it crosses the river.
That angle is θ. VBS will give us the velocity the boat would travel along angle θ. And having calculated angle θ we can determine the
length of vector X.  Knowing the distance, and having previously calculated the velocity, we can then figure out the length of time that
traversing the river would take. And again, knowing the angle θ we can calculate the length of vector D. And knowing that the length of
vector D would be traveled at 3.0 m/s, we can then calculate the time necessary to traverse it. These relationships are 
diagramed in the left side and are illustrated in the displacement vector set. 
→
D
Displacement Vectors Velocity Vectors
→
VWS
→      → → current
R     X VBS →
VBW
  β
θ        θ VBSy VBWy
α
Same angle γ
VBSx VWSx
Note: Cells highlighted in blue are input areas and may be
changed. Cells highlighted in yellow are results and should
not be altered.         VBWx
Displacement Vectors Velocity Vectors
→ Distance directly across River  → Velocity of Water with respect to the Shore
R 1600 m VWS =  0.80 m/s
→ Distance across river at angle θ to straight across → Velocity of Boat with respect to the Water
X 1618 m VBW = 1.50 m/s
→ Distance to be covered on foot → Velocity of Boat with respect to the Shore
D 243 m VBS = 1.39 m/s    at  -8.6 ° from straight across
Steps:
             
Given γ in degrees: 66.8     The boat is pointed 23.2 ° Upstream
→
Calculate VBS: The sum of the two known vectors will give us
→ → → the vector that the boat has relative to the shore.
VBS = VBW + VWS We work this by extracting the x and y components
of the known vectors.     
Calculate VBWx:  Calculate VBWx:
cos γ =  sin γ = 
VBWx     =        VBW  *   cos(γ) VBWy     =        VBW  *   sin(γ)
VBWx     =  1.50 0.394 m/s VBWy     =  1.50 0.919 m/s
VBWx     =  0.59 m/s VBWy     =  1.38 m/s
VBSx      =     VBWx   +   VWSx
VBSx      =    0.59 -0.80 tan α   =
VBSx      =    -0.21 m/s
1.38
VBS          = tan α   =
= -6.59
-0.21
VBS        =[ ( -0.21 ) 2 + (  1.38 ) 2 ] 1/2
VBS          =  1.39 m/s     atan ( tan α)          =  98.6 Degrees
Theta becomes: Now we have the velocity and angle that the boat will
θ    =     90° - α    = -8.6 Positive is degrees upstream have as it travels across the river. We will use those
degrees Negative is degrees downstream in our displacement calculations.  
Now knowing the angle theta, we can calculate our displacement vector lengths:
→ →
Calculate    X: Calculate  D:
→ R →
X         =
D        = R  *  tan(θ)
cos(θ)
→
→ 1600 m D        = 1600 * -0.152
X         =
0.989 Positve is upstream of directly across, negative is downstream
→ →
X         = 1618 m D          = -243 m
                     
Calculate  T1 : Calculate  T2 :
This is the time the boat takes to cross the river. This is the time spent running down the shore.
T1       = |X|  /  VBS T2       = |D|   /  3.0 m/s
T1   =  1618 / 1.39 T2       = 242.6 / 3.0
T1   =  1161 s T2       = 81 s
Calculate Total Time:
T1  +  T2 =  1241 s
Pointed Upstream (Degrees) T1 T2 T1+T2
-10 1083 383 1466
-5 1071 332 1403
0 1067 284 1351
5 1071 239 1310 There may well be other ways of calculating the
10 1083 195 1278 minimum time by building an equation that relates
15 1104 152 1256 γ and θ, but that is not the method used here.  
20 1135 109 1244 After trying a few numbers to get a ballpark range
25 1177 65 1242 we constructed a graph using this workbook's
30 1232 21 1253 algorithms to determine total time vs. the angle
35 1302 26 1328 that the boat was pointed upstream.   
40 1392 76 1468
45 1508 131 1639
50 1659 193 1852
Pointed Upstream (Degrees) T1 T2 T1+T2
20 1135.123 108.582 1243.705
20.2 1136.574 106.858 1243.432
20.4 1138.042 105.133 1243.176 Then we further refined the graph by zooming in to
20.6 1139.529 103.407 1242.936 the range where we knew the answer would be. 
20.8 1141.033 101.681 1242.714          
21 1142.555 99.954 1242.508 The ultimate answer is to point the boat 23.2 
21.2 1144.095 98.226 1242.32 degrees upstream of straight across.   
21.4 1145.653 96.496 1242.149 This grounds the boat on the opposite shore
21.6 1147.229 94.766 1241.995 227 m downstream of straight across. Then  
21.8 1148.824 93.035 1241.859 Friday runs the remaining distance for a total
22 1150.437 91.303 1241.74 elapsed time of 1241 s.    
22.2 1152.069 89.569 1241.638
22.4 1153.719 87.834 1241.554 The workbook has been saved with the correct angle
22.6 1155.389 86.098 1241.487 to arrive at the final answer.    
22.8 1157.077 84.361 1241.438
23 1158.784 82.623 1241.407
23.2     1160.511 80.883 1241.394
23.4 1162.257 79.141 1241.398
23.6 1164.022 77.398 1241.421
23.8 1165.807 75.654 1241.461
24 1167.612 73.908 1241.52
24.2 1169.437 72.160 1241.597
24.4 1171.281 70.411 1241.692
24.6 1173.146 68.660 1241.806
24.8 1175.031 66.907 1241.938
25 1176.936 65.152 1242.089
* The Physics book used a different name for the agent.  I used "Friday" in honor of the great sci-fi writer Robert Heinlein
  whose story "Friday" was one of my favorites.  Friday in Heinlein's book was an agent extraordinaire. Where better to invoke
  her presence than in a problem such as this one?