4
Motion in Two Dimensions
CHAPTER OUTLINE ANSWERS TO QUESTIONS
4.1 The Position, Velocity, and
Acceleration Vectors Q4.1 Yes. An object moving in uniform circular motion moves at a
4.2 Two-Dimensional Motion
constant speed, but changes its direction of motion. An object
with Constant Acceleration
4.3 Projectile Motion cannot accelerate if its velocity is constant.
4.4 Uniform Circular Motion
4.5 Tangential and Radial Q4.2 No, you cannot determine the instantaneous velocity. Yes, you
Acceleration
can determine the average velocity. The points could be widely
4.6 Relative Velocity and
Relative Acceleration separated. In this case, you can only determine the average
velocity, which is
∆x
v=
∆t
.
Q4.3 (a) v (b) v v
v v
a a a a a
v v
a a
a v a v
Q4.4 (a) 10 "i m s (b) −9.80 "j m s 2
Q4.5 The easiest way to approach this problem is to determine acceleration first, velocity second and
Vertical: In free flight, a y = − g . At the top of a projectile’s trajectory, v y = 0. Using this, the
finally position.
maximum height can be found using v 2fy = viy2 + 2 a y y f − yi .
d i
Horizontal: a x = 0 , so v x is always the same. To find the horizontal position at maximum
can be found using v fy = viy + a y t . The horizontal position is x f = vix t .
height, one needs the flight time, t. Using the vertical information found previously, the flight time
If air resistance is taken into account, then the acceleration in both the x and y-directions
would have an additional term due to the drag.
Q4.6 A parabola.
79
,80 Motion in Two Dimensions
Q4.7 The balls will be closest together as the second ball is thrown. Yes, the first ball will always be
moving faster, since its flight time is larger, and thus the vertical component of the velocity is larger.
The time interval will be one second. No, since the vertical component of the motion determines the
flight time.
Q4.8 The ball will have the greater speed. Both the rock and the ball will have the same vertical
component of the velocity, but the ball will have the additional horizontal component.
Q4.9 (a) yes (b) no (c) no (d) yes (e) no
Q4.10 Straight up. Throwing the ball any other direction than straight up will give a nonzero speed at the
top of the trajectory.
Q4.11 No. The projectile with the larger vertical component of the initial velocity will be in the air longer.
Q4.12 The projectile is in free fall. Its vertical component of acceleration is the downward acceleration of
gravity. Its horizontal component of acceleration is zero.
Q4.13 (a) no (b) yes (c) yes (d) no
Q4.14 60°. The projection angle appears in the expression for horizontal range in the function sin 2 θ . This
function is the same for 30° and 60°.
Q4.15 The optimal angle would be less than 45°. The longer the projectile is in the air, the more that air
resistance will change the components of the velocity. Since the vertical component of the motion
determines the flight time, an angle less than 45° would increase range.
Q4.16 The projectile on the moon would have both the larger range and the greater altitude. Apollo
astronauts performed the experiment with golf balls.
Q4.17 Gravity only changes the vertical component of motion. Since both the coin and the ball are falling
from the same height with the same vertical component of the initial velocity, they must hit the floor
at the same time.
Q4.18 (a) no (b) yes
In the second case, the particle is continuously changing the direction of its velocity vector.
Q4.19 The racing car rounds the turn at a constant speed of 90 miles per hour.
Q4.20 The acceleration cannot be zero because the pendulum does not remain at rest at the end of the arc.
Q4.21 (a) The velocity is not constant because the object is constantly changing the direction of its
motion.
(b) The acceleration is not constant because the acceleration always points towards the center of
the circle. The magnitude of the acceleration is constant, but not the direction.
Q4.22 (a) straight ahead (b) in a circle or straight ahead
, Chapter 4 81
Q4.23 v v Q4.24
a a v a v r r
v r r
a a r
a a
a
v a
a a aa
v v v
v v
Q4.25 The unit vectors r" and θ" are in different directions at different points in the xy plane. At a location
along the x-axis, for example, r" = "i and θ" = "j, but at a point on the y-axis, r" = "j and θ" = − i" . The unit
vector "i is equal everywhere, and "j is also uniform.
Q4.26 The wrench will hit at the base of the mast. If air resistance is a factor, it will hit slightly leeward of
the base of the mast, displaced in the direction in which air is moving relative to the deck. If the boat
is scudding before the wind, for example, the wrench’s impact point can be in front of the mast.
Q4.27 (a) The ball would move straight up and down as observed by the passenger. The ball would
move in a parabolic trajectory as seen by the ground observer.
(b) Both the passenger and the ground observer would see the ball move in a parabolic
trajectory, although the two observed paths would not be the same.
Q4.28 (a) g downward (b) g downward
The horizontal component of the motion does not affect the vertical acceleration.
SOLUTIONS TO PROBLEMS
Section 4.1 The Position, Velocity, and Acceleration Vectors
a f a f
−3 600
P4.1 xm ym
0
−3 000 0
−1 270 1 270
−4 270 m −2 330 m
(a) Net displacement = x 2 + y 2
= 4.87 km at 28.6° S of W
FIG. P4.1
Average speed =
b20.0 m sga180 sf + b25.0 m sga120 sf + b30.0 m sga60.0 sf =
180 s + 120 s + 60.0 s
(b) 23.3 m s
4.87 × 10 3 m
(c) Average velocity = = 13.5 m s along R
360 s
, 82 Motion in Two Dimensions
P4.2 (a) r = 18.0t "i + 4.00t − 4.90t 2 "j
e j
(b) v= b18.0 m sg"i + 4.00 m s − e9.80 m s jt "j 2
(c) a= e−9.80 m s j "j 2
(d) ra3.00 sf = a54.0 mf "i − a32.1 mf "j
(e) v 3.00 s = 18.0 m s "i − 25.4 m s "j
a f b g b g
(f) a 3.00 s =
a f e−9.80 m s j "j 2
*P4.3 The sun projects onto the ground the x-component of her velocity:
5.00 m s cos −60.0° = 2.50 m s .
a f
P4.4 (a) From x = −5.00 sin ω t , the x-component of velocity is
−5.00 sin ω t = −5.00ω cos ω t
vx =
dx
=
d FG IJ b g
dt dt H K
and a x = = +5.00ω 2 sin ω t
dv x
dt
similarly, v y =
FG d IJ b4.00 − 5.00 cos ω tg = 0 + 5.00ω sin ω t
H dt K
and a y =
FG d IJ b5.00ω sin ω tg = 5.00ω 2
cos ω t .
H dt K
At t = 0 , v = −5.00ω cos 0 "i + 5.00ω sin 0 "j = e5.00ω "i + 0"jj m s
and a = 5.00ω 2 sin 0 i" + 5.00ω 2 cos 0 "j = e0 i" + 5.00ω "jj m s .
2 2
(b) a4.00 mf"j + a5.00 mfe− sin ω t "i − cos ω t "jj
r = x "i + y"j =
v= a5.00 mfω − cos ω t "i + sin ω t "j
a= a5.00 mfω sin ω t i" + cos ω t "j
2
(c) The object moves in a circle of radius 5.00 m centered at 0 , 4.00 m . a f
Motion in Two Dimensions
CHAPTER OUTLINE ANSWERS TO QUESTIONS
4.1 The Position, Velocity, and
Acceleration Vectors Q4.1 Yes. An object moving in uniform circular motion moves at a
4.2 Two-Dimensional Motion
constant speed, but changes its direction of motion. An object
with Constant Acceleration
4.3 Projectile Motion cannot accelerate if its velocity is constant.
4.4 Uniform Circular Motion
4.5 Tangential and Radial Q4.2 No, you cannot determine the instantaneous velocity. Yes, you
Acceleration
can determine the average velocity. The points could be widely
4.6 Relative Velocity and
Relative Acceleration separated. In this case, you can only determine the average
velocity, which is
∆x
v=
∆t
.
Q4.3 (a) v (b) v v
v v
a a a a a
v v
a a
a v a v
Q4.4 (a) 10 "i m s (b) −9.80 "j m s 2
Q4.5 The easiest way to approach this problem is to determine acceleration first, velocity second and
Vertical: In free flight, a y = − g . At the top of a projectile’s trajectory, v y = 0. Using this, the
finally position.
maximum height can be found using v 2fy = viy2 + 2 a y y f − yi .
d i
Horizontal: a x = 0 , so v x is always the same. To find the horizontal position at maximum
can be found using v fy = viy + a y t . The horizontal position is x f = vix t .
height, one needs the flight time, t. Using the vertical information found previously, the flight time
If air resistance is taken into account, then the acceleration in both the x and y-directions
would have an additional term due to the drag.
Q4.6 A parabola.
79
,80 Motion in Two Dimensions
Q4.7 The balls will be closest together as the second ball is thrown. Yes, the first ball will always be
moving faster, since its flight time is larger, and thus the vertical component of the velocity is larger.
The time interval will be one second. No, since the vertical component of the motion determines the
flight time.
Q4.8 The ball will have the greater speed. Both the rock and the ball will have the same vertical
component of the velocity, but the ball will have the additional horizontal component.
Q4.9 (a) yes (b) no (c) no (d) yes (e) no
Q4.10 Straight up. Throwing the ball any other direction than straight up will give a nonzero speed at the
top of the trajectory.
Q4.11 No. The projectile with the larger vertical component of the initial velocity will be in the air longer.
Q4.12 The projectile is in free fall. Its vertical component of acceleration is the downward acceleration of
gravity. Its horizontal component of acceleration is zero.
Q4.13 (a) no (b) yes (c) yes (d) no
Q4.14 60°. The projection angle appears in the expression for horizontal range in the function sin 2 θ . This
function is the same for 30° and 60°.
Q4.15 The optimal angle would be less than 45°. The longer the projectile is in the air, the more that air
resistance will change the components of the velocity. Since the vertical component of the motion
determines the flight time, an angle less than 45° would increase range.
Q4.16 The projectile on the moon would have both the larger range and the greater altitude. Apollo
astronauts performed the experiment with golf balls.
Q4.17 Gravity only changes the vertical component of motion. Since both the coin and the ball are falling
from the same height with the same vertical component of the initial velocity, they must hit the floor
at the same time.
Q4.18 (a) no (b) yes
In the second case, the particle is continuously changing the direction of its velocity vector.
Q4.19 The racing car rounds the turn at a constant speed of 90 miles per hour.
Q4.20 The acceleration cannot be zero because the pendulum does not remain at rest at the end of the arc.
Q4.21 (a) The velocity is not constant because the object is constantly changing the direction of its
motion.
(b) The acceleration is not constant because the acceleration always points towards the center of
the circle. The magnitude of the acceleration is constant, but not the direction.
Q4.22 (a) straight ahead (b) in a circle or straight ahead
, Chapter 4 81
Q4.23 v v Q4.24
a a v a v r r
v r r
a a r
a a
a
v a
a a aa
v v v
v v
Q4.25 The unit vectors r" and θ" are in different directions at different points in the xy plane. At a location
along the x-axis, for example, r" = "i and θ" = "j, but at a point on the y-axis, r" = "j and θ" = − i" . The unit
vector "i is equal everywhere, and "j is also uniform.
Q4.26 The wrench will hit at the base of the mast. If air resistance is a factor, it will hit slightly leeward of
the base of the mast, displaced in the direction in which air is moving relative to the deck. If the boat
is scudding before the wind, for example, the wrench’s impact point can be in front of the mast.
Q4.27 (a) The ball would move straight up and down as observed by the passenger. The ball would
move in a parabolic trajectory as seen by the ground observer.
(b) Both the passenger and the ground observer would see the ball move in a parabolic
trajectory, although the two observed paths would not be the same.
Q4.28 (a) g downward (b) g downward
The horizontal component of the motion does not affect the vertical acceleration.
SOLUTIONS TO PROBLEMS
Section 4.1 The Position, Velocity, and Acceleration Vectors
a f a f
−3 600
P4.1 xm ym
0
−3 000 0
−1 270 1 270
−4 270 m −2 330 m
(a) Net displacement = x 2 + y 2
= 4.87 km at 28.6° S of W
FIG. P4.1
Average speed =
b20.0 m sga180 sf + b25.0 m sga120 sf + b30.0 m sga60.0 sf =
180 s + 120 s + 60.0 s
(b) 23.3 m s
4.87 × 10 3 m
(c) Average velocity = = 13.5 m s along R
360 s
, 82 Motion in Two Dimensions
P4.2 (a) r = 18.0t "i + 4.00t − 4.90t 2 "j
e j
(b) v= b18.0 m sg"i + 4.00 m s − e9.80 m s jt "j 2
(c) a= e−9.80 m s j "j 2
(d) ra3.00 sf = a54.0 mf "i − a32.1 mf "j
(e) v 3.00 s = 18.0 m s "i − 25.4 m s "j
a f b g b g
(f) a 3.00 s =
a f e−9.80 m s j "j 2
*P4.3 The sun projects onto the ground the x-component of her velocity:
5.00 m s cos −60.0° = 2.50 m s .
a f
P4.4 (a) From x = −5.00 sin ω t , the x-component of velocity is
−5.00 sin ω t = −5.00ω cos ω t
vx =
dx
=
d FG IJ b g
dt dt H K
and a x = = +5.00ω 2 sin ω t
dv x
dt
similarly, v y =
FG d IJ b4.00 − 5.00 cos ω tg = 0 + 5.00ω sin ω t
H dt K
and a y =
FG d IJ b5.00ω sin ω tg = 5.00ω 2
cos ω t .
H dt K
At t = 0 , v = −5.00ω cos 0 "i + 5.00ω sin 0 "j = e5.00ω "i + 0"jj m s
and a = 5.00ω 2 sin 0 i" + 5.00ω 2 cos 0 "j = e0 i" + 5.00ω "jj m s .
2 2
(b) a4.00 mf"j + a5.00 mfe− sin ω t "i − cos ω t "jj
r = x "i + y"j =
v= a5.00 mfω − cos ω t "i + sin ω t "j
a= a5.00 mfω sin ω t i" + cos ω t "j
2
(c) The object moves in a circle of radius 5.00 m centered at 0 , 4.00 m . a f
