# Answers to Relative Velocity Problems

Mechanics

Nov 14, 2013 (4 years and 6 months ago)

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1
.

A jet airliner moving initially at 300 mi/h due east enters a region where the wind is blowing at 100
mi/h in a direction 30.0° north of east. What is the new velocity of the aircraft relative to the ground?

(390 mi/h at 7.37° N of E relative to the
ground)

The velocity of the plane relative to the ground is the vector sum of the velocity of the plane relative to
the air and the velocity of the air relative to the ground, or
PG PA AG
 
v v v
.

The components of this velocity are

PG
300 m ih 100 m ih cos30.0 387 m ih
east
   
v

and

PG
0 100 m ih sin30.0 50.0 m ih
north
   
v

Thus,

2 2
PG
PG PG
2 2
387 50.0 m ih 390 m ih
east north
v
    
v v

and

PG
1 1
PG
50.0
tan tan 7.37
387
north
east

 
 
 
 
   
 
 
 
 
v
v

The plane moves at
390 m ih at 7.37 N of E relative to the g
round

Two Dimensional Kinematics

Relative Velocity

Problems

Name: _______________________________

2.

A boat moves through the water of a river at 10 m/s relative to the water, regar
dless of the boat’s
direction. If the water in the river is flowing at 1.5 m/s, how long does it take the boat to make a round
trip consisting of a 300
-
m displacement downstream followed by a 300
-
m displacement upstream?

We use the following notation:

BS

v

velocity of boat relative to the shore

BW

v

velocity of boat relative to the water,

and

W S

v

velocity of water relative to the shore.

If we take downstream as the positive direction,
then
W S
1.5 m s

v

for both parts of the trip. Also,
BW
10 m s
 
v

while going downstream and
BW
10 m s
 
v

for the upstream part of the trip.

The velocity of the boat relative to the shore is given by
BW W S
BS
 
v v v

While going downstream,
BS
10 m s 1.5 m s
v
 

and the time to go 300 m downstream is

300 m
26 s
10+1.5 m s
down
t
 

When going upstream,
BS
10 m s 1.5 m s 8.5 m s
v
  

and the time required to move 300 m
upstream is
300 m
35 s
8.5 m s
up
t

 

The time for the round trip is

26 35 s 61 s
down up
t t t
    

3
.

A river flows due east at 1.50 m/s. A boat crosses the river from the south shore to the north shore by
maintaining a constant velocity of 10.0 m/s due north relative to the water. (a) What is
the velocity of
the boat relative to the shore? (b) If the river is 300 m wide, how far downstream has the boat moved by
the time it reaches the north shore?

BW
10 m s

v
, directed northward, is the velocity of the boat relative to the
w
ater.

W S
1.5 m s

v
, directed eastward, is the velocity of the water relative to shore.

BS
v

is the velocity of the boat relative to shore, and directed at an angle of

,
relative to the northward direction as shown.

BS BW W S
 
v v v

The northward component of
BS
v

is

BS BW
cos 10 m s
v v

 

(1)

The eastward component is

BS W S
sin 1.5 m s
v v

 

(2)

(a)

Dividing equation (2) by equation (1) gives

1 1
W S
BW
1.50
tan tan 8.53
10.0
v
v

 
 
 
   
 
 
 
 

Fro
m equation (1),

BS
10 m s
10.1 m s
cos8.53
v
 

Therefore,
BS
10.1 m s at 8.53 E of N
 
v

(b)

The time to cross the river is
BS
300 m 300 m
30.0 s
cos 10.0 m s
t
v

  

and the downstream drift of the boat
during this crossing is

BS
sin 1.50 m s 30.0 s 45.0 m
drift v t

  

4.

A rowboat crosses a river with a velocity of 3.30 mi/h at an angle 62.5° north of west relative to the
water. The river is 0.505 mi wide and carries an eastward current of 1.25 mi/h. How far upstream is the
boat when it reaches the opposite shore?

BW

v

velocity of boat relative to the water,

W S

v

velocity of water relative to the shore

and
BS

v

velocity of boat relative to the shore.

BS BW W S
 
v v v

as shown in the
diagram.

The northward (that is, cross
-
stream) component of
BS
v

is

BS BW
sin62.5 0 3.30 m ih sin62.5 0 2.93 m ih
north
v
      
v

The time required to cross the stream is then
0.505 m i
0.173 h
2.93 m ih
t
 

The eastward (that is, downstream) component of
BS
v

is

BW W S
BS
cos62.5
east
v v
  
v

3.30 m ih cos62.5 1.25 m ih 0.274 m ih
   

Since the last result is negative, it is seen that the boat moves upstream as it crosses the river. The
distance it moves upstream is

2
BS
5280 ft
0.274 m ih 0.173 h 4.72 10 m i 249 ft
1 m i
east
d t

 
    
 
 
v

(249 ft)

5
.

How long does it take an automobile traveling in the left lane of a highway at 60.0 km/h to overtake
(become even with) another car that is traveling in the right lane at 40.0 km/h when the cars’ front
bumpers are initially 100 m apart?

Choose the posi
tive direction to be the direction of each car’s motion relative to Earth. The velocity of
the faster car relative to the slower car is given by
FS
FE ES
 
v v v
, where
FE
60.0 km h
 
v

is the
velocity of the faster car relative to Eart
h and
ES SE
40.0 km h
   
v v

is the velocity of Earth relative
to the slower car.

Thus,
FS
60.0 km h 40.0 km h 20.0 km h
   
v

and the time required for the faster car to move 100
m (0.100 km) closer to the slower car is

3
FS
0.100 km 3600 s
5.00 10 h 18.0 s
20.0 km h 1 h
d
t
v

 
    
 
 

6.

A science student is riding on a flatcar of a train traveling along a straight horizontal track at a constant
speed of 10.0 m/s. The student throws a ball along a path that she judges to make an initial angle of
60.0° with the horizontal and to be in line
with the track. The student’s professor, who is standing on
the ground nearby, observes the ball to rise vertically. How high does the ball rise?

(15.3

m)

(7.
23
x

10
3

m, 1.68
x

10
3

m)

BC

v

the velocity of the ball relative to the car

CE

v

velocity of the car relative to Earth
10 m s

BE

v

the velocity of the ball relative to Earth

These velocities are related by the equation
BE CE
BC
 
v v v

as illustrated in the diagram.

Considering the horizontal components, we see that

BE CE
cos60.0
v v
 

or
CE
BC
10.0 m s
20.0m s
cos60.0 cos60.0
v
v
  
 

From the vertical components, the initial velocity of the ball relative to Earth is
BE BC
sin60.0 17.3 m s
v v
  

Using

2 2
0
2
y y y
v v a y
  
, with
0
y
v

when the ball is at maximum height, we find

2
2
2
0
BE
2
m ax
0
17.3 m s
0
15.3 m
2 2
2 9.80 m s
y
y
v
v
y
a g

    

as the maximum height the ball rises.

7
.

Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed
relative to the water. One paddles directly upstream (and moves upstream), whereas the other paddles
directly downstream. With downstream as the positive dir
ection, an observer on shore determines the
velocities of the two canoes to be

1.2 m/s and +2.9 m/s, respectively. (a) What is the speed of the water
relative to the shore? (b) What is the speed of each canoe relative to the water?

(0.85 m/s, 2.1 m/s)

The velocity of a canoe relative to the shore is given by
CS CW W S
 
v v v
, where
CW
v

is the velocity of
the canoe relative to the water and
W S
v

is the velocity of the water relative to shore.

App
lied to the canoe moving upstream, this gives

CW W S
1.2 m s
v v
   

(1)

and for the canoe going downstream

CW W S
2.9 m s
v v
   

(2)

(a)

Adding equations (1) and (2) gives

W S
2 1.7 m s
v

, so
W S
0.85 m s
v

(b) Sub
tracting (1) from (2) yields

CW
2 4.1 m s
v

, or
CW
2.1 m s
v

8
.

A sailboat is heading directly north at a speed of 20 knots (1 knot = 0.514 m/s). The wind is blowing
towards the east with a speed of 17 knots. Determine the magnitude and direction of the wind velocity
as measured on the boat. What is the component of th
e wind velocity in the direction parallel to the
motion of the boat? (See Problem 4.54 for an explanation of how a sailboat can move “into the wind.”)

(26 knots at 50 ° south of east, 20 knots to the south
)

We are given that:

BE
20 knots due north

v

(velocity of boat relative to Earth)

and

W E
17 knots due east

v

(velocity of wind relative to Earth)

The velocity of the wind relative to the boat is

W B W E EB
 
v v v

where
EB BE
20 knots south
  
v v
is the velocity of Earth relative

to the boat. The vector diagram above

Since the vector triangle is a 90° triangle, we find the magnitude of
W B
v

to be

2 2
2 2
W B W E EB
17 knots 20 knots 26 knots
v v v
    

and the direction is given by

1 1
EB
W E
20 knots
tan tan 50
17 knots
v
v

 
 
 
   
 
 
 
 

Thus,
W B
26 knots at 50° south of east

v

From the vector diagram above, the component of this velocity parallel to the motion of the boat (that
is, parallel to a north
-
south line) is seen to be
EB BE
20 knots south
 
v v