6.
5.
4.
1. Darren is canoeing across a river flowing at 3 km/h. He heads out directly across the river; Darren can canoe at a rate of
5 km/h in still water and is moving at this rate to the opposite bank. a) Determine Darren’s resultant velocity. [5.83 km/h] b) If the river is 1.8 km wide at this point, how far downstream will Darren land? [1.08 m downstream] c) How long does it take Darren to cross the river. [0.36 hours]
2. A pilot heads on a course of N25 o W and the airspeed of the plane is 150 km/h. (The airspeed is the speed that the plane would have if there were no wind blowing; i.e. the speed of the plane in still air.) A southwest wind is blowing at 60 km/h. (A south west wind is blowing from the south west to the northeast.) Determine the resultant velocity of the plane. [176.96 km/h N 6.5 ° W]
3. A river is 2 km wide and flows at 6 km/h. A motor boat that has a speed of 20 km/h in still water heads out from one bank perpendicular to the current. A marina lies directly across the river on the opposite bank. a) b)
How far downstream from the marina will the boat reach the other bank? [0.6 km]
How long will it take? [6 min]
An airplane is headed north with a constant velocity of 450 km/h. The plane encounters a west wind blowing at
100 km/h. a) b)
How far will the plane travel in 3 h?
What is the direction of the plane? [N13 o E]
[1383 km]
A light plane is travelling at 175 km/h on a heading of N8 o E in a 40 km/h wind from N80 o E. Determine the plane’s ground velocity (the resultant velocity) [167 km/h, N5 o W]
Fiona heads straight out across a stream flowing at 2 km/h. She can row at 3.5 km/h in still water. a) Determine her resultant velocity. [4.03 km/h, 60.3° to the shore]
7.
8.
9. b) Suppose Fiona had to land on the bank directly opposite her starting position. Determine the angle that she needs to steer and her speed. [55° to the shore, 2.87 km/h]
Gordon is flying his Cessna at an airspeed of 165 km/h. He has set the course as S25 o E. A south wind is blowing at 55 km/h. Determine the ground speed and the resultant direction of the plane. [216.1 km/h]
Haggar is out in his biplane. To reach his destination on time he needs to be on a course of N63 o W at a ground speed of
185 km/h. He measures the wind speed to be 68 km/h from N30 o E. Determine his airspeed and the course he should set to reach his destination on time. [ The airspeed of the plane must be 193.73 km/h and the course set by
Haggar should be N42.48
o W]
Imogen is kayaking across a river flowing at 4 km/h. She wants to land up at a point directly across from her starting point and she can kayak at the rate of 5.5 km/h in still water. [Imogen should head at an angle of 43.34
o to the upstream bank.
Imogen’s velocity across the river is 3.77 km/h. i) At what angle to the upstream bank should Imogen head if she is to be successful in reaching her destination. [ 43.34
o ] ii) How long does it take her to cross the river if it is 2.5 km wide at this point. [39 minutes]

1. Darren is canoeing across a river flowing at 3 km/h. He heads out directly across the river; Darren can canoe at a rate of 5 km/h in still water and is moving at this rate to the opposite bank.
(i)
Determine Darren’s resultant velocity.
(ii) If the river is 1.8 km wide at this point, how far downstream will Darren land?
(iii) How long does it take Darren to cross the river?
Solution:
(i) Darren’s resultant velocity is the vector sum of the velocity of the canoe and the velocity of the current.
5 km/h
3 km/h

As the triangle is a right-angled triangle:
 v
2 
5
2 
3
2
 v

5 .
83 km / h tan
 
3
5
 
30 .
96 o
Darren’s resultant velocity is 5.83 km/h at an angle of 59.04
o to the bank. i.e, the angle with the bank is (90 – 30.96) o .
(ii) The calculation for the distance travelled downstream requires a diagram showing the displacements not the velocities. This triangle is similar to the triangle used to add the velocities.
1.8km
30.96
o x km d km tan 30 .
96

1 .
x
8 x = 1.08 km
Darren lands 1.08 km downstream.
(iii) The triangle representing the velocities and the triangle representing the displacements are similar. Therefore the ratio of the corresponding pairs of sides is constant: t
 d

1 .
8

1 .
08
3 5 .
83 5
Therefore it is not necessary to calculate d to find the time taken. t = 0.36 hour or 21.6 minutes.

7.
2. A pilot heads on a course of N25 o W and the airspeed of the plane is 150 km/h. (The airspeed is the speed that the plane would have if there were no wind blowing; i.e. the speed of the plane in still air.) A southwest wind is blowing at
60 km/h. (A south west wind is blowing from the south west to the northeast.) Determine the resultant velocity of the plane.
Solution:

 v
2 v



150 2 
60 2 
2 ( 150 )( 60 ) cos( 110 )
179 .
60 km / h sin


150

 sin 110
179
51 .
70 o
.
60
The resultant velocity of the plane is
179.60 km/h in a direction N6.70
o W.The resultant velocity is called the ground velocity or sometimes ground speed of the plane.
6. Fiona heads straight out across a stream flowing at 2 km/h. She can row at 3.5 km/h in still water. Determine her resultant velocity.
 v
2 
2
2 
3 .
5
2
 v

4 .
03 km / h
tan
 
2
3 .
5
 
29 .
74 o
Therefore, Fiona’s resultant velocity is 4.03 km/h at
60.26
o to the bank
Gordon is flying his Cessna at an airspeed of 165 km/h. He has set the course as S25 o E. A south wind is blowing at 55 km/h. Determine the ground speed and the resultant direction of the plane. v

R v

2 
165
2 
55
2

117 .
48 km / h

2 ( 165 )( 55 ) cos( 25 o
)
The ground speed is 117.48 km/h sin


55 sin 25
117 .
48
 sin θ =
55 sin 25
117 .
48

11 .
41 o
The heading of the plane is S 36.45
o E i.e, The angle is found by 11.41
o + 25 o = 36.45
o

8.
Haggar is out in his biplane. To reach his destination on time he needs to be on a course of N63 o W at a ground speed of 185 km/h. He measures the wind speed to be 68 km/h from N30 o E. Determine his airspeed and the course he should set to reach his destination on time.
 v a
2 
185
2 
68
2 
2 ( 185 )( 68 ) cos( 87 o
)
 v a

193 .
73 km / h sin


185

 sin 87
193 .
73
72 .
48 o
The airspeed of the plane must be 193.73 km/h and the course set by Haggar should be
N42.48
o W if he is to reach his destination on time.
9. Imogen is kayaking across a river flowing at 4 km/h. She wants to land up at a point directly across from her starting point and she can kayak at the rate of 5.5 km/h in still water. a) At what angle to the upstream bank should Imogen head if she is to be successful in reaching her b) destination.
How long does it take her to cross the river if it is 2.5 km wide at this point.
(a)
(b) v

R
2 
5 .
5
2 
4
2
 v
R

3 .
77 km / h
Imogen’s velocity across the river is 3.77 km/h.
 t

2 .
5
3 .
77 t

0 .
66 h or 39.79mins
Imogen took 39 mins and 47seconds to cross the river. sin
 
4
5 .
5
 
46 .
66 o
Imogen should head at an angle of 43.34
o to the upstream bank.