Two Dimensional Kinematics
Relative Velocity Problems
1.
Name: _______________________________
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 vPG  vPA  vA G .
The components of this velocity are
v 
east
v 
north
PG
and
PG
Thus, vPG 
and
 300 m ih  100 m ih  cos30.0  387 m ih
 0  100 m ih  sin30.0  50.0 m ih
v 
PG
2
east

 vPG
 
 v 

 v
PG
  tan1 
PG
north
east

2
north

 3872   50.02 m ih  390 m ih

  tan1  50.0   7.37

 387 

The plane moves at 390 m ih at7.37 N ofE relative to the ground
2.
A boat moves through the water of a river at 10 m/s relative to the water, regardless 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:
vBS  velocity of boat relative to the shore
vBW  velocity of boat relative to the water,
and
vW S  velocity of water relative to the shore.
If we take downstream as the positive direction, then vW S   1.5 m s for both parts of the trip. Also,
vBW   10 m s while going downstream and vBW   10 m s for the upstream part of the trip.
The velocity of the boat relative to the shore is given by vBS  vBW  vW S
While going downstream, vBS  10 m s 1.5 m s and the time to go 300 m downstream is
300 m
tdow n 
 26 s
 10+1.5 m s
When going upstream, vBS   10 m s 1.5 m s   8.5 m s and the time required to move 300 m
upstream is tup 
 300 m
 35 s
 8.5 m s
The time for the round trip is t tdown  tup   26  35 s 61 s
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?
vBW  10 m s, directed northward, is the velocity of the boat relative to the
water.
vW S  1.5 m s, directed eastward, is the velocity of the water relative to shore.
v BS is the velocity of the boat relative to shore, and directed at an angle of ,
relative to the northward direction as shown.
vBS  vBW  vW S
The northward component of v BS is vBS cos  vBW  10 m s
(1)
vBS sin  vW S  1.5 m s
(2)
The eastward component is
(a) Dividing equation (2) by equation (1) gives
 vW S 
50 
1  1.
  tan 
  8.53
v
10.
 0
 BW 
  tan1 
From equation (1),
vBS 
10 m s
 10.1 m s
cos8.53
Therefore, vBS  10.1 m s at8.53 E ofN
(b) The time to cross the river is t
300 m
300 m

 30.0 s and the downstream drift of the boat
vBS cos 10.0 m s
during this crossing is
drift  vBS sin  t 1.50 m s  30.0 s  45.0 m
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?
vBW  velocity of boat relative to the water,
vW S  velocity of water relative to the shore
and vBS  velocity of boat relative to the shore.
vBS  vBW  vW S as shown in the diagram.
The northward (that is, cross-stream) component of v BS is
 vBS north   vBW  sin62.5  0   3.30 m ih  sin62.5  0  2.93 m ih
The time required to cross the stream is then t
0.505 m i
 0.173 h
2.93 m ih
The eastward (that is, downstream) component of v BS is
v 
BS east
  vBW  cos62.5  vW S
   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
 5280 ft
d   vBS east t  0.274 m ih   0.173 h    4.72 102 m i 
  249 ft
 1m i 
(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 positive 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 vFS  vFE  vES , where vFE   60.0 km h is the
velocity of the faster car relative to Earth and vES   vSE   40.0 km h is the velocity of Earth relative
to the slower car.
Thus, vFS   60.0 km h  40.0 km h   20.0 km h and the time required for the faster car to move 100
m (0.100 km) closer to the slower car is
t
d
0.100 km
 3600 s

 5.00 103 h 
  18.0 s
vFS 20.0 km h
 1h 
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 103 m, 1.68 x 103 m)
v BC  the velocity of the ball relative to the car
v C E  velocity of the car relative to Earth  10 m s
vBE  the velocity of the ball relative to Earth
These velocities are related by the equation vBE  vBC  vCE
as illustrated in the diagram.
Considering the horizontal components, we see that
vBE cos60.0  vCE or vBC 
10.0 m s
vCE

 20.0m s
cos60.0 cos60.0
From the vertical components, the initial velocity of the ball relative to Earth is
vBE  vBC sin 60.0  17.3 m s
Using vy2  v02y  2ay  y , with vy  0 when the ball is at maximum height, we find
 ym ax 
0  v02y
2ay
2
17.3 m s  15.3 m
0  vBE


2  g 2 9.80 m s2 
as the maximum height the ball rises.
2
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 direction, 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 vCS  vCW  vW S , where vCW is the velocity of
the canoe relative to the water and vW S is the velocity of the water relative to shore.
Applied to the canoe moving upstream, this gives
 1.2 m s   vCW  vW S
(1)
and for the canoe going downstream
2.9 m s  vCW  vW S
(a) Adding equations (1) and (2) gives
2vW S  1.7 m s, so vW S  0.85 m s
(b) Subtracting (1) from (2) yields
2vCW  4.1 m s , or vCW  2.1 m s
(2)
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 the 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:
vBE  20 knots due north (velocity of boat relative to Earth)
and
vW E  17 knots due east (velocity of wind relative to Earth)
The velocity of the wind relative to the boat is
vW B  vW E  vEB
where vEB  vBE  20 knots south is the velocity of Earth relative to the boat. The vector diagram above
shows this vector addition.
Since the vector triangle is a 90° triangle, we find the magnitude of v W B to be
2
vW B  vW2 E  vEB

2
2
17 knots   20 knots  26 knots
and the direction is given by
 vEB 
1  20 knots
  tan 
  50
 17 knots
 vW E 
  tan1 
Thus, vW B  26 knots at50° south ofeast
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 vEB  vBE  20 knots south