Cutnell_9th_problems_ch_1_thru_10_converted

and
can be
determined from the data in the tables and the appropriate equations of kinematics.
SOLUTION
(a
) To determine the time that the package is in the air, we will use Equation 3.5b
and the data in the y-direction data table. Solving this quadratic equation for the time yields
We discard the first solution, since it is a negative value and, hence, unrealistic. The displacement can
be found using
, the data in the -direction data table, and Equation 3.5a:



(b The angle that the velocity of the package makes with respect to the ground is given by
)
. Since there is no acceleration in the direction
,
is the same
as
, so that
. Equation 3.3b can be employed with the -direction data to
find
:
Therefore,
where the minus sign indicates that the angle is
.
REASONING
(a) The drawing shows the initial velocity
of the package when it is released. The initial speed of the package is
. The component of its displacement along the ground is labeled as . The data for the x direction are
indicated in the data table below.
x-Direction Data
?
Since only two variables are known, it is not possible to determine from the data in this table. A value
for a third variable is needed. We know that the time of flight is the same for both the and motions, so
let's now look at the data in the direction.
y-Direction Data

?

Note that the displacement of the package points from its initial position toward the ground, so its
value is negative, i.e.,

. The data in this table, along with the appropriate equation of kinematics, can be

used to find the time of flight . This value for can, in turn, be used in conjunction with the

direction data

to determine .

(b) The drawing at the right shows the velocity of the package just before impact. The angle that the
velocity makes

with respect to the ground can be found from the inverse tangent function as
. Once the time
has been found in part (a), the values of
and
can be determined from the data in the tables and the
appropriate equations of kinematics.
SOLUTION
(a)
To determine the time that the package is in the air, we will use Equation 3.5b
and the
data in the y-direction data table. Solving this quadratic equation for the time yields
We discard the first solution, since it is a negative value and, hence, unrealistic. The displacement can
be found using
, the data in the -direction data table, and Equation 3.5a:
(b) The angle that the velocity of the package makes with respect to the ground is given by
.
Since there is no acceleration in the direction
,

is the same as
, so that
. Equation 3.3b can be employed with the -direction data to find
:
Therefore,
where the minus sign indicates that the angle is
.
*38.
Multiple-Concept Example 4 deals with a situation similar to that presented here. A marble is thrown
horizontally with a speed of
from the top of a building. When it strikes the ground, the marble has a velocity
that makes an angle of
with the horizontal. From what height above the ground was the marble thrown?

*39.

Review Conceptual Example 5 before beginning this problem. You are traveling in a convertible with
the top down. The car is moving at a constant velocity of

, due east along flat ground. You throw a tomato

straight upward at a speed of

. How far has the car moved when you get a chance to catch the tomato?
Answer:

56 m

*40.
See Multiple-Concept Example 9 for the basic idea behind problems such as this. A diver springs
upward from a diving board. At the instant she contacts the water, her speed is
, and her body is extended at an angle of
with respect to the horizontal surface of the water. At this instant her vertical displacement is
, where
downward is the negative direction. Determine her initial velocity, both magnitude and direction.
*41.
A soccer player kicks the ball toward a goal that is 16.8 m in front of him. The ball leaves his foot at a
speed of and an angle of
above the ground. Find the speed of the ball when the goalie catches it in front of
the net.
Answer:
*42.
In the javelin throw at a track-and-field event, the javelin is launched at a speed of
at an angle of
above the horizontal. As the javelin travels upward, its velocity points above the horizontal at an angle
that decreases as time passes. How much time is required for the angle to be reduced from
at launch to
?
*43.
An airplane is flying with a velocity of
at an angle of
with the horizontal, as the drawing
shows. When the altitude of the plane is 2.4 km, a flare is released from the plane. The flare hits the
target on the ground. What is the angle ?

Answer:
REASONING The angle can be found from
(1)
where is the horizontal displacement of the flare. Since
, it follows that
. The
flight time is determined by the vertical motion. In particular, the time t can be found from Equation
3.5b. Once the time is known, can be calculated.
SOLUTION From Equation 3.5b, assuming upward is the positive direction, we have
which can be rearranged to give the following equation that is quadratic in t:
Using
and
and suppressing the units, we obtain the quadratic equation

Using the quadratic formula, we obtain

. Therefore, we find that

Equation 1 then gives

*44.

A child operating a radio-controlled model car on a dock accidentally steers it off the edge. The car's
displacement 1.1 s after leaving the dock has a magnitude of 7.0 m. What is the car's speed at the
instant it drives off the edge of the dock?

*45.

After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 51.0 m
horizontally from the end of the ramp. His velocity, just before landing, is

and points in a direction

below the

horizontal. Neglecting air resistance and any lift he experiences while airborne, find his initial
velocity (magnitude and direction) when he left the end of the ramp. Express the direction as an angle
relative to the horizontal.

Answer:

,

above the horizontal

*46.

Stones are thrown horizontally with the same velocity from the tops of two different buildings. One
stone lands twice as far from the base of the building from which it was thrown as does the other
stone. Find the ratio of the height of the taller building to the height of the shorter building.

**47.

The drawing shows an exaggerated view of a rifle that has been “sighted in” for a 91.4-meter target. If
the muzzle speed of the bullet is

, what are the two possible angles and between the rifle barrel and

the horizontal such that the bullet will hit the target? One of these angles is so large that it is never
used in target shooting.

Problem 47

( Hint: The following trigonometric identity may be useful:

.)

Answer:

and

REASONING AND SOLUTION In the absence of air resistance, the bullet exhibits projectile motion.
The component of the motion has zero acceleration while the component of the motion is subject to
the acceleration due to gravity. The horizontal distance traveled by the bullet is given by Equation 3.5a
(with

):

with equal to the time required for the bullet to reach the target. The time can be found by considering
the vertical motion. From Equation 3.3b,

When the bullet reaches the target,

. Assuming that up and to the right are the positive directions, we

have

Using the fact that
, we have
Thus, we find that

and

Therefore,

**48.

A projectile is launched from ground level at an angle of

above the horizontal. It returns to ground level.

To what value should the launch angle be adjusted, without changing the launch speed, so that the
range doubles?

**49.

From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of

, parallel to the

ground. As the drawing shows, the bullet puts a hole in a window of another building and hits the wall
that faces the window. Using the data in the drawing, determine the distances D and H, which locate the
point where the gun was fired. Assume that the bullet does not slow down as it passes through the
window.

Answer:

REASONING Since the horizontal motion is not accelerated, we know that the component of the
velocity remains constant at 340 m/s. Thus, we can use Equation 3.5a (with

) to determine the time that the bullet

spends in the building before it is embedded in the wall. Since we know the vertical displacement of
the bullet after it enters the building, we can use the flight time in the building and Equation 3.5b to
find the component of the velocity of the bullet as it enters the window. Then, Equation 3.6b can be
used (with

) to determine the

vertical displacement of the bullet as it passes between the buildings. We can determine the distance by
adding the magnitude of to the vertical distance of 0.50 m within the building.

Once we know the vertical displacement of the bullet as it passes between the buildings, we can
determine the time required for the bullet to reach the window using Equation 3.4b. Since the motion
in the direction is not accelerated, the distance D can then be found from

.

SOLUTION Assuming that the direction to the right is positive, we find that the time that the bullet
spends in the building is (according to Equation 3.5a)

The vertical displacement of the bullet after it enters the building is, taking down as the negative
direction, equal to

. Therefore, the vertical component of the velocity of the bullet as it passes through the window is,
from Equation 3.5b,

The vertical displacement of the bullet as it travels between the buildings is (according to Equation
3.6b with

)

Therefore, the distance is

The time for the bullet to reach the window, according to Equation 3.4b, is

Hence, the distance is given by

**50. In the annual battle of the dorms, students gather on the roofs of Jackson and Walton dorms to
launch water balloons at each other with slingshots. The horizontal distance between the buildings is
35.0 m, and the heights of the Jackson and Walton buildings are, respectively, 15.0 m and 22.0 m.

Ignore air resistance.

(a) The first balloon launched by the Jackson team hits Walton dorm 2.0 s after launch, striking it
halfway between the ground and the roof. Find the direction of the balloon's initial velocity. Give
your answer as an angle measured above the horizontal.

(b) A second balloon launched at the same angle hits the edge of Walton's roof. Find the initial speed
of this second balloon.

**51. Two cannons are mounted as shown in the drawing and rigged to fire simultaneously. They are
used in a circus act in which two clowns serve as human cannonballs. The clowns are fired toward
each other and collide at a height of 1.00

m above the muzzles of the cannons. Clown A is launched at a

angle, with a speed of

. The

horizontal separation between the clowns as they leave the cannons is 6.00 m. Find the launch speed
and the

launch angle

for clown B.

Answer:
,
Section 3.4 Relative Velocity
52. In a marathon race Chad is out in front, running due north at a speed of
. John is 95 m behind him, running
due north at a speed of
. How long does it take for John to pass Chad?
53.

A swimmer, capable of swimming at a speed of
in still water (i.e., the swimmer can swim with a speed
of
relative to the water), starts to swim directly across a 2.8-km-wide river. However, the current is
, and it carries the swimmer downstream.
(a) How long does it take the swimmer to cross the river?
Answer:
REASONING The velocity
of the swimmer relative to the ground is the vector sum of the
velocity
of the swimmer relative to the water and the velocity
of the water relative to the ground
as shown at the right:
.
The component of
that is parallel to the width of the river determines how fast the swimmer is moving
across the river; this parallel component is
. The time for the swimmer to cross the river is equal to the
width of the river divided by the magnitude of this velocity component.
The component of
that is parallel to the direction of the current determines how far the swimmer is
carried down stream; this component is
. Since the motion occurs with constant velocity, the distance
that the swimmer is carried downstream while crossing the river is equal to the magnitude of
multiplied by the time it takes for the swimmer to cross the river.

SOLUTION
(a) The time for the swimmer to cross the river is
(b) The distance that the swimmer is carried downstream while crossing the river is
(b) How far downstream will the swimmer be upon reaching the other side of the river?
Answer:
REASONING The velocity
of the swimmer relative to the ground is the vector sum of the
velocity
of the swimmer relative to the water and the velocity
of the water relative to the ground
as shown at the right:
.

The component of
that is parallel to the width of the river determines how fast the swimmer is moving
across the river; this parallel component is
. The time for the swimmer to cross the river is equal to the
width of the river divided by the magnitude of this velocity component.
The component of
that is parallel to the direction of the current determines how far the swimmer is
carried down stream; this component is
. Since the motion occurs with constant velocity, the distance
that the swimmer is carried downstream while crossing the river is equal to the magnitude of
multiplied by the time it takes for the swimmer to cross the river.
SOLUTION
(a) The time for the swimmer to cross the river is

(b) The distance that the swimmer is carried downstream while crossing the river is
REASONING The velocity
of the swimmer relative to the ground is the vector sum of the velocity
of the swimmer relative to the water and the velocity
of the water relative to the ground as shown at the
right:
.
The component of
that is parallel to the width of the river determines how fast the swimmer is moving across the
river; this parallel component is
. The time for the swimmer to cross the river is equal to the width of the river
divided by the magnitude of this velocity component.
The component of
that is parallel to the direction of the current determines how far the swimmer is carried down
stream; this component is
. Since the motion occurs with constant velocity, the distance that the swimmer is
carried downstream while crossing the river is equal to the magnitude of
multiplied by the time it takes for the
swimmer to cross the river.
SOLUTION
(a) The time for the swimmer to cross the river is



(b) The distance that the swimmer is carried downstream while crossing the river is

54.

Two friends, Barbara and Neil, are out rollerblading. With respect to the ground, Barbara is skating
due south at a speed of

. Neil is in front of her. With respect to the ground, Neil is skating due west at a speed of

.

Find Neil's velocity (magnitude and direction relative to due west), as seen by Barbara.

55. A police officer is driving due north at a constant speed of

relative to the ground when she notices a truck on

an east–west highway ahead of her, driving west at high speed. She finds that the truck's speed relative
to her car is (about 110 mph).

(a) Sketch the vector triangle that shows how the truck's velocity relative to the ground is related to
the police car's velocity relative to the ground and to the truck's velocity relative to the police car. The
sketch need not be to scale, but the velocity vectors should be oriented correctly and bear the
appropriate labels.

Answer:

The answer is a drawing.

(b) What is the truck's speed, relative to the ground?

Answer:

56. At some airports there are speed ramps to help passengers get from one place to another. A speed
ramp is a moving conveyor belt on which you can either stand or walk. Suppose a speed ramp has a
length of 105 m and is moving at a speed of

relative to the ground. In addition, suppose you can cover this distance in 75 s when walking on the
ground. If you walk at the same rate with respect to the speed ramp that you walk on the ground, how
long does it take for you to travel the 105 m using the speed ramp?

57. You are in a hot-air balloon that, relative to the ground, has a velocity of

in a direction due east. You see a
hawk moving directly away from the balloon in a direction due north. The speed of the hawk relative
to you is
. What are the magnitude and direction of the hawk's velocity relative to the ground? Express the
directional angle relative to due east.
Answer:
,
north of east
58.
On a pleasure cruise a boat is traveling relative to the water at a speed of
due south. Relative to the boat,
a passenger walks toward the back of the boat at a speed of
.
(a) What are the magnitude and direction of the passenger's velocity relative to the water?
(b) How long does it take for the passenger to walk a distance of 27 m on the boat?
(c) How long does it take for the passenger to cover a distance of 27 m on the water?
59.
Two passenger trains are passing each other on adjacent tracks. Train A is moving east with a speed
of
,
and train B is traveling west with a speed of
.
(a) What is the velocity (magnitude and direction) of train A as seen by the passengers in train B?
Answer:
due east
REASONING The velocity
of train A relative to train B is the vector sum of the velocity

of train A relative to the ground and the velocity
of the ground relative to train B, as indicated by
Equation 3.7:
. The values of
and
are given in the statement of the problem.
We must also make use of the fact that
.
SOLUTION
(a) Taking east as the positive direction, the velocity of A relative to B is, according to Equation 3.7,
The positive sign indicates that the direction of
is
.
(b) Similarly, the velocity of B relative to A is
The negative sign indicates that the direction of
is
.
(b) What is the velocity (magnitude and direction) of train B as seen by the passengers in train A?
Answer:
due west



REASONING The velocity
of train A relative to train B is the vector sum of the velocity
of train A relative to the ground and the velocity
of the ground relative to train B, as indicated by
Equation 3.7:

. The values of
and
are given in the statement of the problem.
We must also make use of the fact that
.
SOLUTION
(a) Taking east as the positive direction, the velocity of A relative to B is, according to Equation 3.7,
The positive sign indicates that the direction of
is
.
(b) Similarly, the velocity of B relative to A is
The negative sign indicates that the direction of
is
.
REASONING The velocity
of train A relative to train B is the vector sum of the velocity
of train A
relative to the ground and the velocity
of the ground relative to train B, as indicated by Equation 3.7:
. The values of
and
are given in the statement of the problem. We must also make use
of the fact that
.
SOLUTION
(a) Taking east as the positive direction, the velocity of A relative to B is, according to Equation 3.7,

The positive sign indicates that the direction of
is
.
(b) Similarly, the velocity of B relative to A is
The negative sign indicates that the direction of
is
.
60.
The captain of a plane wishes to proceed due west. The cruising speed of the plane is
relative to the air.
A weather report indicates that a 38.0-m/s wind is blowing from the south to the north. In what
direction, measured with respect to due west, should the pilot head the plane?
*61.
A person looking out the window of a stationary train notices that raindrops are falling vertically
down at a speed of
relative to the ground. When the train moves at a constant velocity, the raindrops make an angle of
when they move past the window, as the drawing shows. How fast is the train moving?
Answer:
*62.
A ferryboat is traveling in a direction
north of east with a speed of
relative to the water. A
passenger is walking with a velocity of
due east relative to the boat. What is the velocity (magnitude and
direction) of the passenger with respect to the water? Determine the directional angle relative to due
east.
*63.

Mario, a hockey player, is skating due south at a speed of
relative to the ice. A teammate passes the
puck to him. The puck has a speed of
and is moving in a direction of
west of south, relative to the ice.
What are the magnitude and direction (relative to due south) of the puck's velocity, as observed by
Mario?
Answer:
,
west of south
REASONING The velocity
of the puck relative to Mario is the vector sum of the velocity
of the puck
relative to the ice and the velocity
of the ice relative to Mario as indicated by Equation 3.7:
.
The values of
and
are given in the statement of the problem. In order to use the data, we must make use of the



fact that
, with the result that
.
SOLUTION The first two rows of the following table give the east/west and north/south components
of the vectors and
. The third row gives the components of their resultant
. Due east and due north
have been taken as positive.
Vector
East/West Component
North/South Component

0

Now that the components of
are known, the Pythagorean theorem can be used to find the magnitude.
The direction of
is found from
**64. A jetliner can fly 6.00 hours on a full load of fuel. Without any wind it flies at a speed of
. The plane
is to make a round-trip by heading due west for a certain distance, turning around, and then heading
due east for the return trip. During the entire flight, however, the plane encounters a 57.8-m/s wind
from the jet stream, which blows from west to east. What is the maximum distance that the plane can
travel due west and just be able to return home?
**65.
Two boats are heading away from shore. Boat 1 heads due north at a speed of
relative to the shore.
Relative to boat 1, boat 2 is moving
north of east at a speed of
. A passenger on boat 2 walks due
east across the deck at a speed of
relative to boat 2. What is the speed of the passenger relative to the shore?
Answer:
REASONING The relative velocities in this problem are:
= velocity of the Passenger relative to the Shore
= velocity of the Passenger relative to Boat 2 (1.20 m/s, due east)
= velocity of Boat 2 relative to the Shore
= velocity of Boat 2 relative to Boat 1 (1.60 m/s, at
north of east)

= velocity of Boat 1 relative to the Shore (3.00 m/s, due north)
The velocity
of the passenger relative to the shore is related to
and
by (see the method of subscripting
discussed in Section 3.4):
But
, the velocity of Boat 2 relative to the shore, is related to
and
by

Substituting this expression for
into the first equation yields
This vector sum is shown in the diagram. We will determine the magnitude of
from the equation above by using
the method of scalar components.
SOLUTION The table below lists the scalar components of the four vectors in the drawing.
Vector
x Component
y Component

0 m/s
0 m/s
The magnitude of
can be found by applying the Pythagorean theorem to its and components:
Copyright © 2012 John Wiley & Sons, Inc. Al rights reserved.
Problems
Section 4.3 Newton's Second Law of Motion
1. An airplane has a mass of
and takes off under the influence of a constant net force of
.
What is the net force that acts on the plane's 78-kg pilot?
Answer:

93 N

2.
A boat has a mass of 6800 kg. Its engines generate a drive force of 4100 N due west, while the wind
exerts a force of 800 N due east and the water exerts a resistive force of 1200 N due east. What are the
magnitude and direction of the boat's acceleration?
3.
Two horizontal forces,
and
, are acting on a box, but only
is shown in the drawing.
can point
either to the right or to the left. The box moves only along the x axis. There is no friction between the
box and the surface. Suppose that
and the mass of the box is 3.0 kg. Find the magnitude and direction of
when the acceleration of the box is (a)
, (b)
, and (c)
.



Answer:
(a)
(b)
(c)
4. In the amusement park ride known as Magic Mountain Superman, powerful magnets accelerate a
car and its riders from rest to
(about
) in a time of 7.0 s. The combined mass of the car and riders is
.
Find the average net force exerted on the car and riders by the magnets.
5.
A person in a kayak starts paddling, and it accelerates from 0 to
in a distance of 0.41 m. If the
combined mass of the person and the kayak is 73 kg, what is the magnitude of the net force acting on
the kayak?
Answer:

32 N

REASONING The magnitude
of the net force acting on the kayak is given by Newton's second law as
(Equation 4.1), where is the combined mass of the person and kayak, and is their acceleration. Since
the initial and final velocities,
and , and the displacement are known, we can employ one of the equations of
kinematics from Chapter 2 to find the acceleration.
SOLUTION Solving Equation 2.9
from the equations of kinematics for the acceleration, we have
Substituting this result into Newton's second law gives
6. Scientists are experimenting with a kind of gun that may eventually be used to fire payloads
directly into orbit. In one test, this gun accelerates a 5.0-kg projectile from rest to a speed of
. The net force accelerating the
projectile is
. How much time is required for the projectile to come up to speed?
7.
A 1580-kg car is traveling with a speed of
. What is the magnitude of the horizontal net force
that is required to bring the car to a halt in a distance of 50.0 m?
Answer:

3560 N

REASONING AND SOLUTION The acceleration required is

Newton's second law then gives the magnitude of the net force as

8.

The space probe Deep Space 1 was launched on October 24, 1998. Its mass was 474 kg. The goal of
the mission was to test a new kind of engine called an ion propulsion drive. This engine generated
only a weak thrust, but it could do so over long periods of time with the consumption of only small
amounts of fuel. The mission was spectacularly successful. At a thrust of 56 mN how many days were
required for the probe to attain a velocity of

, assuming that the probe started from rest and that the mass remained nearly constant?

*9.

Two forces

and

are applied to an object whose mass is 8.0 kg. The larger force is

. When both

forces point due east, the object's acceleration has a magnitude of
. However, when
points due east
and
points due west, the acceleration is
, due east. Find (a) the magnitude of
and (b) the
magnitude of
.

Answer:
(a) 3.6 N
(b) 0.40 N
REASONING Let due east be chosen as the positive direction. Then, when both forces point due east,
Newton's second law gives
(4.1)
where
. When
points due east and
points due west, Newton's second law gives
(4.2)
where
. These two equations can be used to find the magnitude of each force.
SOLUTION
(a) Adding Equations 4.1 and 4.2 gives
(b) Subtracting Equation 4.2 from Equation 4.1 gives
*10.
An electron is a subatomic particle
that is subject to electric forces. An electron
moving in the
direction accelerates from an initial velocity of
to a final velocity of
while traveling a distance of 0.038 m. The electron's acceleration is due to two electric forces
parallel to the x axis:
, and
, which points in the

direction. Find the magnitudes
of (a) the net force acting on the electron and (b) the electric force
.
Sect io n 4.4 T he Vect o r Nat ure o f Newt o n's Seco nd Law o f Mo t io n
Sect io n 4.5 Newt o n's T hird Law o f Mo t io n
11. Only two forces act on an object
, as in the drawing. Find the magnitude and direction (relative to
the x axis) of the acceleration of the object.
Problem 11
Answer:

12. At an instant when a soccer ball is in contact with the foot of a player kicking it, the horizontal or
x component of the ball's acceleration is
and the vertical or y component of its acceleration is
. The ball's mass is
0.43 kg. What is the magnitude of the net force acting on the soccer ball at this instant?
13.
A rocket of mass
is in flight. Its thrust is directed at an angle of
above the
horizontal and has a magnitude of
. Find the magnitude and direction of the rocket's acceleration. Give
the direction as an angle above the horizontal.
Answer:
REASONING To determine the acceleration we will use Newton's second law
. Two forces act
on the rocket, the thrust and the rocket's weight , which is

. Both of these forces must be considered when determining
the net force
. The direction of the acceleration is the same as the direction of the net force.
SOLUTION In constructing the free-body diagram for the rocket we choose upward and to the right
as the positive directions. The free-body diagram is as follows:
The component of the net force is
The component of the net force is
The magnitudes of the net force and of the acceleration are
The direction of the acceleration is the same as the direction of the net force. Thus, it is directed
above the horizontal at an angle of
14.
A billiard ball strikes and rebounds from the cushion of a pool table perpendicularly. The mass of the
ball is 0.38
kg. The ball approaches the cushion with a velocity of
and rebounds with a velocity of
. The
ball remains in contact with the cushion for a time of
. What is the average net force (magnitude and
direction) exerted on the ball by the cushion?

15. When a parachute opens, the air exerts a large drag force on it. This upward force is initially
greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky

diver is 915 N and the drag force has a magnitude of 1027 N. The mass of the sky diver is 93.4 kg.
What are the magnitude and direction of his acceleration?

Answer:

16.

Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades
and the ice.

The mass of the man is 82 kg, and the mass of the woman is 48 kg. The woman pushes on the man
with a force of 45

N due east. Determine the acceleration (magnitude and direction) of (a) the man and (b) the woman.

*17.

A space probe has two engines. Each generates the same amount of force when fired, and the
directions of these forces can be independently adjusted. When the engines are fired simultaneously
and each applies its force in the same direction, the probe, starting from rest, takes 28 s to travel a
certain distance. How long does it take to travel the same distance, again starting from rest, if the
engines are fired simultaneously and the forces that they apply to the probe are perpendicular?

Answer:

33 s

**18. At a time when mining asteroids has become feasible, astronauts have connected a line between
their 3500-kg space tug and a 6200-kg asteroid. Using their tug's engine, they pull on the asteroid with
a force of 490 N. Initially the tug and the asteroid are at rest, 450 m apart. How much time does it take
for the tug and the asteroid to meet?
**19.
A 325-kg boat is sailing
north of east at a speed of
. Thirty seconds later, it is sailing
north of east at a speed of
. During this time, three forces act on the boat: a 31.0-N force
directed
north of east (due to an auxiliary engine), a 23.0-N force directed
south of west
(resistance due to the water), and
(due to the wind). Find the magnitude and direction of the force
.
Express the direction as an angle with respect to due east.
Answer:
REASONING We first determine the acceleration of the boat. Then, using Newton's second law, we
can find the net force
that acts on the boat. Since two of the three forces are known, we can solve for the
unknown force
once the net force
is known.
SOLUTION Let the direction due east be the positive direction and the direction due north be the
positive direction. The and components of the initial velocity of the boat are then

Thirty seconds later, the and velocity components of the boat are

Therefore, according to Equations 3.3a and 3.3b, the and components of the acceleration of the boat
are Thus, the and components of the net force that act on the boat are