M03 YOUN7066 13 ISM c03 kho tài liệu bách khoa

3

MOTION IN TWO OR THREE DIMENSIONS

3.1.

IDENTIFY and SET UP: Use Eq. (3.2), in component form.
Δx x2 − x1 5.3 m − 1.1 m
EXECUTE: (a) (vav ) x =
=
=
= 1.4 m/s
Δt t2 − t1
3.0 s − 0

(vav ) y =

Δy y2 − y1 −0.5 m − 3.4 m
=
=
= −1.3 m/s
Δt
t2 − t1
3.0 s − 0

(b)

tan α =

(vav ) y
(vav ) x

=

−1.3 m/s
= −0.9286
1.4 m/s

α = 360° − 42.9° = 317°
vav = (vav ) 2x + (vav ) 2y

vav = (1.4 m/s) 2 + (−1.3 m/s) 2 = 1.9 m/s
Figure 3.1

G
EVALUATE: Our calculation gives that vav is in the 4th quadrant. This corresponds to increasing x and
3.2.

decreasing y.
G
IDENTIFY: Use Eq. (3.2), written in component form. The distance from the origin is the magnitude of r .
SET UP: At time t1, x1 = y1 = 0.
EXECUTE: (a) x = (vav-x )Δt = (−3.8 m/s)(12.0 s) = −45.6 m and y = (vav-y )Δt = (4.9 m/s)(12.0 s) = 58.8 m.
(b) r = x 2 + y 2 = (−45.6 m) 2 + (58.8 m)2 = 74.4 m.
G
G
EVALUATE: Δr is in the direction of vav . Therefore, Δx is negative since vav-x is negative and Δy is

positive since vav-y is positive.

3.3.

G
(a) IDENTIFY and SET UP: From r we can calculate x and y for any t.
Then use Eq. (3.2), in component form.
G
EXECUTE: r = [4.0 cm + (2.5 cm/s 2 )t 2 ]iˆ + (5.0 cm/s)tˆj
G
At t = 0, r = (4.0 cm) iˆ.
G
At t = 2.0 s, r = (14.0 cm) iˆ + (10.0 cm) ˆj.

Δx 10.0 cm
=
= 5.0 cm/s.
Δt
2.0 s
Δy 10.0 cm
(vav ) y =
=
= 5.0 cm/s.
Δt
2.0 s
(vav ) x =

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3-1

3-2

Chapter 3

vav = (vav ) 2x + (vav ) 2y = 7.1 cm/s
tan α =

(vav ) y
(vav ) x

= 1.00

θ = 45°.

Figure 3.3a

G
EVALUATE: Both x and y increase, so vav is in the 1st quadrant.
G
G
(b) IDENTIFY and SET UP: Calculate r by taking the time derivative of r (t ).
G
G dr
EXECUTE: v =
= ([5.0 cm/s 2 ]t )iˆ + (5.0 cm/s) ˆj
dt
t = 0: vx = 0, v y = 5.0 cm/s; v = 5.0 cm/s and θ = 90°
t = 1.0 s: vx = 5.0 cm/s, v y = 5.0 cm/s; v = 7.1 cm/s and θ = 45°
t = 2.0 s: vx = 10.0 cm/s, v y = 5.0 cm/s; v = 11 cm/s and θ = 27°
(c) The trajectory is a graph of y versus x.

x = 4.0 cm + (2.5 cm/s 2 ) t 2 , y = (5.0 cm/s)t

For values of t between 0 and 2.0 s, calculate x and y and plot y versus x.

Figure 3.3b

3.4.

EVALUATE: The sketch shows that the instantaneous velocity at any t is tangent to the trajectory.
IDENTIFY: Given the position vector of a squirrel, find its velocity components in general, and at a
specific time find its velocity components and the magnitude and direction of its position vector and
velocity.
SET UP: vx = dx/dt and vy = dy/dt; the magnitude of a vector is A = ( Ax2 + Ay2 ).
EXECUTE: (a) Taking the derivatives gives vx (t ) = 0.280 m/s + (0.0720 m/s 2 )t and

v y (t ) = (0.0570 m/s3 )t 2 .

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Motion in Two or Three Dimensions

3-3

(b) Evaluating the position vector at t = 5.00 s gives x = 2.30 m and y = 2.375 m, which gives
r = 3.31 m.
1.425
(c) At t = 5.00 s, vx = +0.64 m/s, v y = 1.425 m/s, which gives v = 1.56 m/s and tan θ =
so the

0.64

3.5.

direction is θ = 65.8o (counterclockwise from +x-axis)
EVALUATE: The acceleration is not constant, so we cannot use the standard kinematics formulas.
IDENTIFY and SET UP: Use Eq. (3.8) in component form to calculate (aav ) x and (aav ) y .
EXECUTE: (a) The velocity vectors at t1 = 0 and t2 = 30.0 s are shown in Figure 3.5a.

Figure 3.5a

Δvx v2 x − v1x −170 m/s − 90 m/s
=
=
= −8.67 m/s 2
Δt
t2 − t1
30.0 s
Δv y v2 y − v1 y 40 m/s − 110 m/s
(aav ) y =
=
=
= −2.33 m/s 2
Δt
30.0 s
t2 − t1

(b) (aav ) x =

(c)

a = (aav ) 2x + ( aav ) 2y = 8.98 m/s 2

tan α =

(aav ) y

=

−2.33 m/s 2

−8.67 m/s 2
α = 15° + 180° = 195°
( aav ) x

= 0.269

Figure 3.5b
EVALUATE: The changes in vx and v y are both in the negative x or y direction, so both components of
G
aav are in the 3rd quadrant.
3.6.

IDENTIFY: Use Eq. (3.8), written in component form.
SET UP: ax = (0.45m/s 2 )cos31.0° = 0.39m/s2 , a y = (0.45m/s2 )sin 31.0° = 0.23m/s 2
EXECUTE: (a) aav-x =

Δv y
Δvx
and vx = 2.6 m/s + (0.39 m/s 2 )(10.0 s) = 6.5 m/s. aav-y =

and
Δt
Δt

v y = −1.8 m/s + (0.23 m/s 2 )(10.0 s) = 0.52 m/s.
⎛ 0.52 ⎞
(b) v = (6.5m/s)2 + (0.52m/s) 2 = 6.52m/s, at an angle of arctan ⎜
⎟ = 4.6° above the horizontal.
⎝ 6.5 ⎠
G
G
(c) The velocity vectors v1 and v2 are sketched in Figure 3.6. The two velocity vectors differ in

magnitude and direction.
G
EVALUATE: v1 is at an angle of 35° below the +x-axis and has magnitude v1 = 3.2 m/s, so v2 > v1 and
G
G
the direction of v2 is rotated counterclockwise from the direction of v1.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


3-4

Chapter 3

Figure 3.6
3.7.

IDENTIFY and SET UP: Use Eqs. (3.4) and (3.12) to find vx , v y , ax , and a y as functions of time. The
G
G
magnitude and direction of r and a can be found once we know their components.
EXECUTE: (a) Calculate x and y for t values in the range 0 to 2.0 s and plot y versus x. The results are
given in Figure 3.7a.

Figure 3.7a

dx
dy
= α vy =
= −2β t
dt
dt
dv y
dv
= −2β
ay = x = 0 ay =
dt
dt
G
G
Thus v = α iˆ − 2β tˆj a = −2β ˆj
(b) vx =

(c) velocity: At t = 2.0 s, vx = 2.4 m/s, v y = −2(1.2 m/s 2 )(2.0 s) = −4.8 m/s

v = vx2 + v 2y = 5.4 m/s

tan α =

vy
vx

=

−4.8 m/s
= −2.00
2.4 m/s

α = −63.4° + 360° = 297°

Figure 3.7b

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Motion in Two or Three Dimensions

3-5

acceleration: At t = 2.0 s, ax = 0, a y = −2 (1.2 m/s 2 ) = −2.4 m/s 2

a = ax2 + a 2y = 2.4 m/s 2
tan β =

ay
ax

=

−2.4 m/s 2
= −∞
0

β = 270°
Figure 3.7c

G
EVALUATE: (d) a has a component ai in the same
G
direction as v , so we know that v is increasing (the bird
G
is speeding up.) a also has a component a⊥
G
G
perpendicular to v , so that the direction of v is
changing; the bird is turning toward the − y -direction

(toward the right)
Figure 3.7d

G
G
v is always tangent to the path; v at t = 2.0 s shown in part (c) is tangent to the path at this t, conforming
G
G
to this general rule. a is constant and in the − y -direction; the direction of v is turning toward the

− y -direction.
3.8.

IDENTIFY: Use the velocity components of a car (given as a function of time) to find the acceleration of
the car as a function of time and to find the magnitude and direction of the car’s velocity and acceleration
at a specific time.
SET UP: a x = dvx /dt and a y = dv y /dt ; the magnitude of a vector is A = ( Ax2 + Ay2 ).
EXECUTE: (a) Taking the derivatives gives a x (t ) = (−0.0360 m/s3 )t and a y (t ) = 0.550 m/s 2 .
(b) Evaluating the velocity components at t = 8.00 s gives vx = 3.848 m/s and v y = 6.40 m/s, which gives

v = 7.47 m/s. The direction is tan θ =

6.40
so θ = 59.0o (counterclockwise from +x-axis).
3.848

(c) Evaluating the acceleration components at t = 8.00 s gives a x = 20.288 m/s 2 and a y = 0.550 m/s 2 ,

which gives a = 0.621 m/s 2 . The angle with the +y axis is given by tan θ =

3.9.

0.288
, so θ = 27.6o. The
0.550

direction is therefore 118o counterclockwise from +x-axis.
EVALUATE: The acceleration is not constant, so we cannot use the standard kinematics formulas.
IDENTIFY: The book moves in projectile motion once it leaves the table top. Its initial velocity is
horizontal.

SET UP: Take the positive y-direction to be upward. Take the origin of coordinates at the initial position
of the book, at the point where it leaves the table top.

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3-6

Chapter 3

x-component:
a x = 0, v0 x = 1.10 m/s,
t = 0.350 s
y-component:
a y = −9.80 m/s 2 ,

v0 y = 0,
t = 0.350 s

Figure 3.9a

Use constant acceleration equations for the x and y components of the motion, with a x = 0 and a y = − g .
EXECUTE: (a) y − y0 = ?

y − y0 = v0 yt + 12 a yt 2 = 0 + 12 ( −9.80 m/s 2 )(0.350 s) 2 = −0.600 m. The table top is 0.600 m above the floor.
(b) x − x0 = ?

x − x0 = v0 xt + 12 axt 2 = (1.10 m/s)(0.350 s) + 0 = 0.385 m.
(c) vx = v0 x + a xt = 1.10 m/s (The x-component of the velocity is constant, since a x = 0.)

v y = v0 y + a yt = 0 + ( −9.80 m/s 2 )(0.350 s) = −3.43 m/s
v = vx2 + v 2y = 3.60 m/s
tan α =

vy
vx

=

−3.43 m/s
= −3.118
1.10 m/s

α = −72.2°
G
Direction of v is 72.2° below the horizontal

Figure 3.9b
(d) The graphs are given in Figure 3.9c.

Figure 3.9c
EVALUATE: In the x-direction, a x = 0 and vx is constant. In the y-direction, a y = −9.80 m/s 2 and v y is

downward and increasing in magnitude since a y and v y are in the same directions. The x and y motions
occur independently, connected only by the time. The time it takes the book to fall 0.600 m is the time it
travels horizontally.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Motion in Two or Three Dimensions
3.10.

3-7

IDENTIFY: The person moves in projectile motion. She must travel 1.75 m horizontally during the time
she falls 9.00 m vertically.
SET UP: Take + y downward. a x = 0, a y = +9.80 m/s 2 . v0 x = v0 , v0 y = 0.
EXECUTE: Time to fall 9.00 m: y − y0 = v0 yt + 12 a yt 2 gives t =

2( y − y0 )
2(9.00 m)
=
= 1.36 s.
ay
9.80 m/s 2

Speed needed to travel 1.75 m horizontally during this time: x − x0 = v0 xt + 12 axt 2 gives
x − x0 1.75 m
=
= 1.29 m/s.
t
1.36 s
EVALUATE: If she increases her initial speed she still takes 1.36 s to reach the level of the ledge, but has
traveled horizontally farther than 1.75 m.
IDENTIFY: Each object moves in projectile motion.
SET UP: Take + y to be downward. For each cricket, a x = 0 and a y = +9.80 m/s 2 . For Chirpy,
v0 = v0 x =

3.11.

v0 x = v0 y = 0. For Milada, v0 x = 0.950 m/s, v0 y = 0.
EXECUTE: Milada’s horizontal component of velocity has no effect on her vertical motion. She also
reaches the ground in 3.50 s. x − x0 = v0 xt + 12 axt 2 = (0.950 m/s)(3.50 s) = 3.32 m

3.12.

EVALUATE: The x and y components of motion are totally separate and are connected only by the fact that
the time is the same for both.
IDENTIFY: The football moves in projectile motion.
SET UP: Let + y be upward. a x = 0, a y = − g . At the highest point in the trajectory, v y = 0.

v0 y

12.0m/s
=
= 1.224 s, which we round to 1.22 s.
g
9.80m/s 2
(b) Different constant acceleration equations give different expressions but the same numerical result:
EXECUTE: (a) v y = v0 y + a yt. The time t is

1
2

gt 2 = 12 v y 0t =

v02y

= 7.35 m.
2g
(c) Regardless of how the algebra is done, the time will be twice that found in part (a), which is
2(1.224 s) = 2.45 s.
(d) a x = 0, so x − x0 = v0 xt = (20.0 m/s)(2.45 s) = 49.0 m.
(e) The graphs are sketched in Figure 3.12.
EVALUATE: When the football returns to its original level, vx = 20.0 m/s and v y = −12.0 m/s.

Figure 3.12

3.13.

IDENTIFY: The car moves in projectile motion. The car travels 21.3 m − 1.80 m = 19.5 m downward
during the time it travels 61.0 m horizontally.
SET UP: Take + y to be downward. a x = 0, a y = +9.80 m/s 2 . v0 x = v0 , v0 y = 0.
EXECUTE: (a) Use the vertical motion to find the time in the air:
2( y − y0 )
2(19.5 m)
=
= 1.995 s
y − y0 = v0 yt + 12 a yt 2 gives t =
ay
9.80 m/s 2

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


3-8

Chapter 3

Then x − x0 = v0 xt + 12 a xt 2 gives v0 = v0 x =

x − x0 61.0 m
=
= 30.6 m/s.
t
1.995 s

(b) vx = 30.6 m/s since a x = 0. v y = v0 y + a yt = −19.6m s. v = vx2 + v 2y = 36.3m s.
3.14.

EVALUATE: We calculate the final velocity by calculating its x and y components.
IDENTIFY: Knowing the maximum reached by the froghopper and its angle of takeoff, we want to find its
takeoff speed and the horizontal distance it travels while in the air.
SET UP: Use coordinates with the origin at the ground and + y upward. a x = 0, a y = − 9.80 m/s 2 . At the

maximum height v y = 0. The constant-acceleration formulas v 2y = v02y + 2a y ( y − y0 ) and
y − y0 = v0 yt + 12 a yt 2 apply.
EXECUTE: (a) v 2y = v02y + 2a y ( y − y0 ) gives

v y = −2a y ( y − y0 ) = −2(−9.80 m/s 2 )(0.587 m) = 3.39 m/s. v0 y = v0 sin θ0 so
v0 =

v0 y
sin θ 0

=

3.39 m/s
= 4.00 m/s.
sin 58.0°

(b) Use the vertical motion to find the time in the air. When the froghopper has returned to the ground,
2v0 y
2(3.39 m/s)
y − y0 = 0. y − y0 = v0 yt + 12 a yt 2 gives t = −
=−
= 0.692 s.
ay
−9.80 m/s 2

Then x − x0 = v0 xt + 12 axt 2 = (v0 cos θ0 )t = (4.00 m/s)(cos 58.0°)(0.692 s) = 1.47 m.
EVALUATE: v y = 0 when t = −
3.15.

v0 y
ay

=−

3.39 m/s
−9.80 m/s 2

= 0.346 s. The total time in the air is twice this.

IDENTIFY: The ball moves with projectile motion with an initial velocity that is horizontal and has
magnitude v0 . The height h of the table and v0 are the same; the acceleration due to gravity changes from

g E = 9.80 m/s 2 on earth to g X on planet X.
SET UP: Let + x be horizontal and in the direction of the initial velocity of the marble and let + y be
upward. v0 x = v0 , v0 y = 0, ax = 0, a y = − g , where g is either g E or g X .
EXECUTE: Use the vertical motion to find the time in the air: y − y0 = − h. y − y0 = v0 yt + 12 a yt 2 gives

t=

2h
2h
. Then x − x0 = v0 xt + 12 axt 2 gives x − x0 = v0 xt = v0
. x − x0 = D on earth and 2.76D on
g
g

Planet X. ( x − x0 ) g = v0 2h , which is constant, so D g E = 2.76 D g X .

3.16.

gE

= 0.131g E = 1.28 m/s 2 .
(2.76)2
EVALUATE: On Planet X the acceleration due to gravity is less, it takes the ball longer to reach the floor
and it travels farther horizontally.
IDENTIFY: The shell moves in projectile motion.
SET UP: Let +x be horizontal, along the direction of the shell’s motion, and let + y be upward. ax = 0,

gX =

a y = −9.80 m/s 2 .
EXECUTE: (a) v0 x = v0 cos α 0 = (50.0 m/s)cos 60.0° = 25.0 m/s,

v0 y = v0 sin α 0 = (50.0 m/s)sin 60.0° = 43.3 m/s.
(b) At the maximum height v y = 0. v y = v0 y + a y t gives t =

v y − v0 y
ay

=

0 − 43.3 m/s
= 4.42 s.
−9.80 m/s 2

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Motion in Two or Three Dimensions

(c) v 2y = v02y + 2 a y ( y − y0 ) gives y − y0 =

v y2 − v02y
2a y

=

3-9

0 − (43.3 m/s) 2
= 95.7 m.
2(−9.80 m/s 2 )

(d) The total time in the air is twice the time to the maximum height, so
x − x0 = v0 xt + 12 axt 2 = (25.0 m/s)(8.84 s) = 221 m.
(e) At the maximum height, vx = v0 x = 40.0 m/s and v y = 0. At all points in the motion, ax = 0 and

a y = −9.80 m/s 2 .
EVALUATE: The equation for the horizontal range R derived in the text is R =

v02 sin 2α 0
. This gives
g

(50.0 m/s) 2 sin(120.0°)
= 221 m, which agrees with our result in part (d).
9.80 m/s 2
IDENTIFY: The baseball moves in projectile motion. In part (c) first calculate the components of the
velocity at this point and then get the resultant velocity from its components.
SET UP: First find the x- and y-components of the initial velocity. Use coordinates where the
+ y -direction is upward, the + x -direction is to the right and the origin is at the point where the baseball

R=

3.17.

leaves the bat.

v0 x = v0 cos α 0 = (30.0 m/s) cos36.9° = 24.0 m/s

v0 y = v0 sin α 0 = (30.0 m/s) sin 36.9° = 18.0 m/s

Figure 3.17a

Use constant acceleration equations for the x and y motions, with ax = 0 and a y = − g .
EXECUTE: (a) y-component (vertical motion):
y − y0 = +10.0 m/s, v0 y = 18.0 m/s, a y = −9.80 m/s 2 , t = ?

y − y0 = v0 y + 12 a yt 2
10.0 m = (18.0 m/s)t − (4.90 m/s 2 )t 2
(4.90 m/s 2 )t 2 − (18.0 m/s)t + 10.0 m = 0
1 ⎡18.0 ± ( −18.0) 2 − 4 (4.90)(10.0) ⎤ s = (1.837 ± 1.154) s
Apply the quadratic formula: t = 9.80
⎣⎢
⎦⎥

The ball is at a height of 10.0 above the point where it left the bat at t1 = 0.683 s and at t2 = 2.99 s. At the
earlier time the ball passes through a height of 10.0 m as its way up and at the later time it passes through
10.0 m on its way down.
(b) vx = v0 x = +24.0 m/s, at all times since a x = 0.
v y = v0 y + a yt
t1 = 0.683 s: v y = +18.0 m/s + ( −9.80 m/s 2 )(0.683 s) = +11.3 m/s. (v y is positive means that the ball is
traveling upward at this point.
t2 = 2.99 s: v y = +18.0 m/s + (−9.80 m/s 2 )(2.99 s) = −11.3 m/s. (v y is negative means that the ball is
traveling downward at this point.)

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

3-10

Chapter 3
(c) vx = v0 x = 24.0 m/s

Solve for v y :
v y = ?, y − y0 = 0 (when ball returns to height where motion started),
a y = −9.80 m/s 2 , v0 y = +18.0 m/s
v 2y = v02y + 2a y ( y − y0 )
v y = −v0 y = −18.0 m/s (negative, since the baseball must be traveling downward at this point)
G
Now solve for the magnitude and direction of v .
v = vx2 + v 2y
v = (24.0 m/s)2 + (−18.0 m/s) 2 = 30.0 m/s
tan α =

vy
vx

=

−18.0 m/s
24.0 m/s

α = −36.9°, 36.9° below the horizontal
Figure 3.17b

The velocity of the ball when it returns to the level where it left the bat has magnitude 30.0 m/s and is
directed at an angle of 36.9° below the horizontal.

EVALUATE: The discussion in parts (a) and (b) explains the significance of two values of t for which
y − y0 = +10.0 m. When the ball returns to its initial height, our results give that its speed is the same as its

3.18.

initial speed and the angle of its velocity below the horizontal is equal to the angle of its initial velocity
above the horizontal; both of these are general results.
IDENTIFY: The shot moves in projectile motion.
SET UP: Let + y be upward.
EXECUTE: (a) If air resistance is to be ignored, the components of acceleration are 0 horizontally and
− g = −9.80 m/s 2 vertically downward.
(b) The x-component of velocity is constant at vx = (12.0 m/s)cos51.0° = 7.55 m/s. The y-component is

v0 y = (12.0 m/s) sin 51.0° = 9.32 m/s at release and

v y = v0 y − gt = (9.32 m/s) − (9.80 m/s)(2.08 s) = −11.06 m/s when the shot hits.
(c) x − x0 = v0 xt = (7.55 m/s)(2.08 s) = 15.7 m.
(d) The initial and final heights are not the same.
(e) With y = 0 and v0 y as found above, Eq. (3.18) gives y0 = 1.81m.
(f) The graphs are sketched in Figure 3.18.
EVALUATE: When the shot returns to its initial height, v y = −9.32 m/s. The shot continues to accelerate

downward as it travels downward 1.81 m to the ground and the magnitude of v y at the ground is larger
than 9.32 m/s.

Figure 3.18

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Motion in Two or Three Dimensions
3.19.

3-11

IDENTIFY: Take the origin of coordinates at the point where the quarter leaves your hand and take
positive y to be upward. The quarter moves in projectile motion, with a x = 0, and a y = − g . It travels

vertically for the time it takes it to travel horizontally 2.1 m.
v0 x = v0 cos α 0 = (6.4 m/s) cos60°
v0 x = 3.20 m/s
v0 y = v0 sin α 0 = (6.4 m/s) sin 60°

v0 y = 5.54 m/s
Figure 3.19
(a) SET UP: Use the horizontal (x-component) of motion to solve for t, the time the quarter travels
through the air:
t = ?, x − x0 = 2.1 m, v0 x = 3.2 m/s, a x = 0

x − x0 = v0 xt + 12 a xt 2 = v0 xt , since a x = 0
EXECUTE: t =

x − x0
2.1 m
=
= 0.656 s
v0 x
3.2 m/s

SET UP: Now find the vertical displacement of the quarter after this time:
y − y0 = ?, a y = −9.80 m/s 2 , v0 y = +5.54 m/s, t = 0.656 s

y − y0 + v0 yt + 12 a yt 2
EXECUTE:

y − y0 = (5.54 m/s)(0.656 s) + 12 (−9.80 m/s 2 )(0.656 s)2 = 3.63 m − 2.11 m = 1.5 m.

(b) SET UP: v y = ?, t = 0.656 s, a y = −9.80 m/s 2 , v0 y = +5.54 m/s v y = v0 y + a yt
EXECUTE: v y = 5.54 m/s + (−9.80 m/s 2 )(0.656 s) = −0.89 m/s.

G
EVALUATE: The minus sign for v y indicates that the y-component of v is downward. At this point the
quarter has passed through the highest point in its path and is on its way down. The horizontal range if it
returned to its original height (it doesn’t!) would be 3.6 m. It reaches its maximum height after traveling
horizontally 1.8 m, so at x − x0 = 2.1 m it is on its way down.
3.20.

IDENTIFY: Use the analysis of Example 3.10.
d
SET UP: From Example 3.10, t =
and ydart = (v0 sin α 0 )t − 12 gt 2 .
v0 cos α 0
EXECUTE: Substituting for t in terms of d in the expression for ydart gives

⎛
⎞
gd
ydart = d ⎜ tan α 0 − 2
⎟.

2
⎜
2v0 cos α 0 ⎟⎠
⎝
Using the given values for d and α 0 to express this as a function of v0 ,
⎛
26.62 m 2 /s 2 ⎞
y = (3.00 m) ⎜ 0.90 −
⎟⎟ .
⎜
v02
⎝
⎠
(a) v0 = 12.0 m/s gives y = 2.14 m.
(b) v0 = 8.0 m/s gives y = 1.45 m.
(c) v0 = 4.0 m/s gives y = −2.29 m. In this case, the dart was fired with so slow a speed that it hit the

ground before traveling the 3-meter horizontal distance.
EVALUATE: For (a) and (b) the trajectory of the dart has the shape shown in Figure 3.26 in the textbook.
For (c) the dart moves in a parabola and returns to the ground before it reaches the x-coordinate of the
monkey.

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3-12
3.21.

Chapter 3

IDENTIFY: Take the origin of coordinates at the roof and let the + y -direction be upward. The rock moves

in projectile motion, with a x = 0 and a y = − g . Apply constant acceleration equations for the x and y
components of the motion.
SET UP:
v0 x = v0 cos α 0 = 25.2 m/s
v0 y = v0 sin α 0 = 16.3 m/s

Figure 3.21a
(a) At the maximum height v y = 0.

a y = −9.80 m/s 2 , v y = 0, v0 y = +16.3 m/s, y − y0 = ?
v 2y = v02y + 2a y ( y − y0 )
EXECUTE:

y − y0 =

v 2y − v02y
2a y

=

0 − (16.3 m/s) 2
2(−9.80 m/s 2 )

= +13.6 m

(b) SET UP: Find the velocity by solving for its x and y components.
vx = v0 x = 25.2 m/s (since a x = 0)

v y = ?, a y = −9.80 m/s 2 , y − y0 = −15.0 m (negative because at the ground the rock is below its initial
position), v0 y = 16.3 m/s
v 2y = v02y + 2a y ( y − y0 )
v y = − v02y + 2a y ( y − y0 ) (v y is negative because at the ground the rock is traveling downward.)
EXECUTE: v y = − (16.3 m/s) 2 + 2(−9.80 m/s 2 )(−15.0 m) = −23.7 m/s

Then v = vx2 + v 2y = (25.2 m/s) 2 + ( −23.7 m/s) 2 = 34.6 m/s.
(c) SET UP: Use the vertical motion (y-component) to find the time the rock is in the air:

t = ?, v y = −23.7 m/s (from part (b)), a y = −9.80 m/s 2 , v0 y = +16.3 m/s
EXECUTE: t =

v y − v0 y
ay

=

−23.7 m/s − 16.3 m/s
−9.80 m/s 2

= +4.08 s

SET UP: Can use this t to calculate the horizontal range:
t = 4.08 s, v0 x = 25.2 m/s, a x = 0, x − x0 = ?
EXECUTE:

x − x0 = v0 xt + 12 a xt 2 = (25.2 m/s)(4.08 s) + 0 = 103 m

(d) Graphs of x versus t, y versus t, vx versus t and v y versus t:

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Motion in Two or Three Dimensions

3-13

Figure 3.21b
EVALUATE: The time it takes the rock to travel vertically to the ground is the time it has to travel
horizontally. With v0 y = +16.3 m/s the time it takes the rock to return to the level of the roof ( y = 0) is

t = 2v0 y /g = 3.33 s. The time in the air is greater than this because the rock travels an additional 15.0 m to
3.22.

the ground.
IDENTIFY: Consider the horizontal and vertical components of the projectile motion. The water travels
45.0 m horizontally in 3.00 s.
SET UP: Let + y be upward. a x = 0, a y = −9.80 m/s 2 . v0 x = v0 cosθ0 , v0 y = v0 sin θ0 .
EXECUTE: (a) x − x0 = v0 xt + 12 axt 2 gives x − x0 = v0 (cos θ 0 )t and cosθ0 =

45.0 m
= 0.600;
(25.0 m/s)(3.00 s)

θ 0 = 53.1°
(b) At the highest point vx = v0 x = (25.0 m/s)cos 53.1° = 15.0 m/s, v y = 0 and v = vx2 + v 2y = 15.0 m/s. At

all points in the motion, a = 9.80 m/s 2 downward.
(c) Find y − y0 when t = 3.00s:

y − y0 = v0 yt + 12 a yt 2 = (25.0 m/s)(sin53.1°)(3.00 s) + 12 (−9.80 m/s 2 )(3.00 s)2 = 15.9 m
vx = v0 x = 15.0 m/s, v y = v0 y + a yt = (25.0 m/s)(sin53.1°) − (9.80m/s 2 )(3.00 s) = −9.41 m/s, and
v = vx2 + v 2y = (15.0 m/s) 2 + (−9.41 m/s) 2 = 17.7 m/s
EVALUATE: The acceleration is the same at all points of the motion. It takes the water
v0 y
20.0 m/s
t=−
=−
= 2.04 s to reach its maximum height. When the water reaches the building it has
ay
−9.80 m/s 2
3.23.

passed its maximum height and its vertical component of velocity is downward.
IDENTIFY and SET UP: The stone moves in projectile motion. Its initial velocity is the same as that of the
balloon. Use constant acceleration equations for the x and y components of its motion. Take + y to be
downward.
EXECUTE: (a) Use the vertical motion of the rock to find the initial height.
t = 6.00 s, v0 y = +20.0 m/s, a y = +9.80 m/s 2 , y − y0 = ?
y − y0 = v0 yt + 12 a yt 2 gives y − y0 = 296 m
(b) In 6.00 s the balloon travels downward a distance y − y0 = (20.0 m/s)(6.00 s) = 120 m. So, its height
above ground when the rock hits is 296 m − 120 m = 176 m.
(c) The horizontal distance the rock travels in 6.00 s is 90.0 m. The vertical component of the distance

between the rock and the basket is 176 m, so the rock is

(176 m) 2 + (90 m) 2 = 198 m from the basket

when it hits the ground.

(d) (i) The basket has no horizontal velocity, so the rock has horizontal velocity 15.0 m/s relative to the
basket. Just before the rock hits the ground, its vertical component of velocity is
v y = v0 y + a yt = 20.0 m/s + (9.80 m/s 2 )(6.00 s) = 78.8 m/s, downward, relative to the ground. The basket is
moving downward at 20.0 m/s, so relative to the basket the rock has a downward component of velocity 58.8 m/s.
(ii) horizontal: 15.0 m/s; vertical: 78.8 m/s
EVALUATE: The rock has a constant horizontal velocity and accelerates downward

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3-14
3.24.

Chapter 3
IDENTIFY: We want to find the acceleration of the inner ear of a dancer, knowing the rate at which she spins.
1.0 s
SET UP: R = 0.070 m. For 3.0 rev/s, the period T (time for one revolution) is T =
= 0.333 s. The
3.0 rev

speed is v = d/T = (2πR)/T, and arad = v 2 /R.
v 2 (2π R/T )2 4π 2 R 4π 2 (0.070 m)
=
=
=
= 25 m/s 2 = 2.5 g .
R
R
(0.333 s) 2

T2
EVALUATE: The acceleration is large and the force on the fluid must be 2.5 times its weight.
IDENTIFY: Apply Eq. (3.30).
SET UP: T = 24 h.
EXECUTE: arad =
3.25.

EXECUTE: (a) arad =

4π 2 (6.38 × 106 m)
((24 h)(3600 s/h)) 2

= 0.034 m/s 2 = 3.4 × 10−3 g .

(b) Solving Eq. (3.30) for the period T with arad = g , T =

4π 2 (6.38 × 106 m)
9.80 m/s 2

EVALUATE: arad is proportional to 1/T 2 , so to increase arad by a factor of

= 5070 s = 1.4 h.

1
3.4 × 10−3

= 294 requires

1
24 h

.
= 1.4 h.
294
294
IDENTIFY: Each blade tip moves in a circle of radius R = 3.40 m and therefore has radial acceleration
that T be multiplied by a factor of

3.26.

arad = v 2 /R.

SET UP: 550 rev/min = 9.17 rev/s, corresponding to a period of T =
EXECUTE: (a) v =
(b) arad =

v2
= 1.13 × 104 m/s 2 = 1.15 × 103 g .
R

EVALUATE: arad =
3.27.

2π R
= 196 m/s.
T

1
= 0.109 s.
9.17 rev/s

4π 2 R

gives the same results for arad as in part (b).
T2
IDENTIFY: For the curved lowest part of the dive, the pilot’s motion is approximately circular. We know
the pilot’s acceleration and the radius of curvature, and from this we want to find the pilot’s speed.
v2
SET UP: arad = 5.5 g = 53.9 m/s 2 . 1 mph = 0.4470 m/s. arad = .
R
v2
, so v = Rarad = (350 m)(53.9 m/s 2 ) = 140 m/s = 310 mph.
R
EVALUATE: This speed is reasonable for the type of plane flown by a test pilot.
IDENTIFY: Each planet moves in a circular orbit and therefore has acceleration arad = v 2 /R.
EXECUTE: arad =

3.28.

SET UP: The radius of the earth’s orbit is r = 1.50 × 1011 m and its orbital period is

T = 365 days = 3.16 × 107 s. For Mercury, r = 5.79 × 1010 m and T = 88.0 days = 7.60 × 106 s.
EXECUTE: (a) v =
(b) arad =

2π r
= 2.98 × 104 m/s
T

v2
= 5.91 × 10−3 m/s 2 .

r

(c) v = 4.79 × 104 m/s, and arad = 3.96 × 10−2 m/s 2 .
EVALUATE: Mercury has a larger orbital velocity and a larger radial acceleration than earth.

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Motion in Two or Three Dimensions
3.29.

3-15

IDENTIFY: Uniform circular motion.
G
dv
G
SET UP: Since the magnitude of v is constant, vtan =
= 0 and the resultant acceleration is equal to
dt
the radial component. At each point in the motion the radial component of the acceleration is directed in
toward the center of the circular path and its magnitude is given by v 2 /R.

v 2 (7.00 m/s) 2
=
= 3.50 m/s 2 , upward.
R
14.0 m
(b) The radial acceleration has the same magnitude as in part (a), but now the direction toward the center of

the circle is downward. The acceleration at this point in the motion is 3.50 m/s 2 , downward.
(c) SET UP: The time to make one rotation is the period T, and the speed v is the distance for one
revolution divided by T.
2π R
2π R 2π (14.0 m)
EXECUTE: v =
so T =
=
= 12.6 s
T
7.00 m/s
v
EVALUATE: The radial acceleration is constant in magnitude since v is constant and is at every point in
G
the motion directed toward the center of the circular path. The acceleration is perpendicular to v and is
G
nonzero because the direction of v changes.
EXECUTE: (a) arad =

3.30.

v2
. The speed in rev/s is
R
1/ T , where T is the period in seconds (time for 1 revolution). The speed v increases with R along the
length of his body but all of him rotates with the same period T.
SET UP: For his head R = 8.84 m and for his feet R = 6.84 m.
IDENTIFY: Each part of his body moves in uniform circular motion, with arad =

EXECUTE: (a) v = Rarad = (8.84 m)(12.5)(9.80 m/s 2 ) = 32.9 m/s

(b) Use arad =

T = 2π

4π 2 R
T2

. Since his head has arad = 12.5 g and R = 8.84 m,

R
8.84m
R 4π 2 (6.84m)
= 2π
=1.688s. Then his feet have arad = 2 =
= 94.8m/s 2 = 9.67 g.
2
arad
12.5(9.80m/s )
T
(1.688s)2

The difference between the acceleration of his head and his feet is 12.5 g − 9.67 g = 2.83 g = 27.7 m/s 2 .
(c)

1
1
=
= 0.592 rev/s = 35.5 rpm
T 1.69 s

EVALUATE: His feet have speed v = Rarad = (6.84 m)(94.8 m/s 2 ) = 25.5 m/s
3.31.

IDENTIFY: Relative velocity problem. The time to walk the length of the moving sidewalk is the length
divided by the velocity of the woman relative to the ground.
SET UP: Let W stand for the woman, G for the ground and S for the sidewalk. Take the positive direction
to be the direction in which the sidewalk is moving.
The velocities are vW/G (woman relative to the ground), vW/S (woman relative to the sidewalk), and vS/G

(sidewalk relative to the ground).
Eq. (3.33) becomes vW/G = vW/S + vS/G .
The time to reach the other end is given by t =

distance traveled relative to ground
vW/G

EXECUTE: (a) vS/G = 1.0 m/s

vW/S = +1.5 m/s
vW/G = vW/S + vS/G = 1.5 m/s + 1.0 m/s = 2.5 m/s.
t=

35.0 m 35.0 m
=
= 14 s.
vW/G
2.5 m/s

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3-16

Chapter 3
(b) vS/G = 1.0 m/s

vW/S = −1.5 m/s
vW/G = vW/S + vS/G = −1.5 m/s + 1.0 m/s = −0.5 m/s. (Since vW/G now is negative, she must get on the
moving sidewalk at the opposite end from in part (a).)
−35.0 m −35.0 m
t=
=
= 70 s.
vW/G
−0.5 m/s

3.32.

EVALUATE: Her speed relative to the ground is much greater in part (a) when she walks with the motion
of the sidewalk.
G
G
IDENTIFY: The relative velocities are vS/F , the velocity of the scooter relative to the flatcar, vS/G , the
G
G
G
G
scooter relative to the ground and vF/G , the flatcar relative to the ground. vS/G = vS/F + vF/G . Carry out the

vector addition by drawing a vector addition diagram.
G
G
G
G
G
SET UP: vS/F = vS/G − vF/G . vF/G is to the right, so −vF/G is to the left.
EXECUTE: In each case the vector addition diagram gives
(a) 5.0 m/s to the right
(b) 16.0 m/s to the left
(c) 13.0 m/s to the left.
EVALUATE: The scooter has the largest speed relative to the ground when it is moving to the right relative
G
G
to the flatcar, since in that case the two velocities vS/F and vF/G are in the same direction and their
3.33.

3.34.

3.35.

magnitudes add.
IDENTIFY: Apply the relative velocity relation.
G
G
SET UP: The relative velocities are vC/E , the canoe relative to the earth, vR/E , the velocity of the river
G
relative to the earth and vC/R , the velocity of the canoe relative to the river.
G
G

G
G
G
G
G
EXECUTE: vC/E = vC/R + vR/E and therefore vC/R = vC/E − vR/E . The velocity components of vC/R are
−0.50 m/s + (0.40 m/s)/ 2, east and (0.40 m/s)/ 2, south, for a velocity relative to the river of 0.36 m/s,
at 52.5° south of west.
EVALUATE: The velocity of the canoe relative to the river has a smaller magnitude than the velocity of
the canoe relative to the earth.
IDENTIFY: Calculate the rower’s speed relative to the shore for each segment of the round trip.
SET UP: The boat’s speed relative to the shore is 6.8 km/h downstream and 1.2 km/h upstream.
EXECUTE: The walker moves a total distance of 3.0 km at a speed of 4.0 km/h, and takes a time of three
fourths of an hour (45.0 min).
1.5 km
1.5 km
The total time the rower takes is
+
= 1.47 h = 88.2 min.
6.8 km/h 1.2 km/h
EVALUATE: It takes the rower longer, even though for half the distance his speed is greater than 4.0 km/h.
The rower spends more time at the slower speed.
IDENTIFY: Relative velocity problem in two dimensions. His motion relative to the earth (time
displacement) depends on his velocity relative to the earth so we must solve for this velocity.
(a) SET UP: View the motion from above.

The velocity vectors in the problem are:
G
vM/E , the velocity of the man relative to the earth
G

vW/E , the velocity of the water relative to the earth
G
vM/W , the velocity of the man relative to the water
The rule for adding these velocities is
G
G
G
vM/E = vM/W + v W/E

Figure 3.35a
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Motion in Two or Three Dimensions

3-17

G
G
The problem tells us that vW/E has magnitude 2.0 m/s and direction due south. It also tells us that vM/W
has magnitude 4.2 m/s and direction due east. The vector addition diagram is then as shown in Figure 3.35b.

This diagram shows the vector addition
G
G
G
vM/E = vM/W + v W/E
G
G

and also has vM/W and v W/E in their
specified directions. Note that the vector
diagram forms a right triangle.
Figure 3.35b
2
2
2
= vM/W
+ vW/E
.
The Pythagorean theorem applied to the vector addition diagram gives vM/E
2
2
EXECUTE: vM/E = vM/W
+ vW/E
= (4.2 m/s) 2 + (2.0 m/s) 2 = 4.7 m/s; tan θ =

vM/W 4.2 m/s
=
= 2.10;
vW/E 2.0 m/s

θ = 65°; or φ = 90° − θ = 25°. The velocity of the man relative to the earth has magnitude 4.7 m/s and
direction 25° S of E.
(b) This requires careful thought. To cross the river the man must travel 800 m due east relative to the
G
earth. The man’s velocity relative to the earth is vM/E . But, from the vector addition diagram the eastward
component of vM/E equals vM/W = 4.2 m/s.
x − x0 800 m
=

= 190 s.
vx
4.2 m/s
G
(c) The southward component of vM/E equals vW/E = 2.0 m/s. Therefore, in the 190 s it takes him to cross

Thus t =

the river, the distance south the man travels relative to the earth is
y − y0 = v yt = (2.0 m/s)(190 s) = 380 m.
EVALUATE: If there were no current he would cross in the same time, (800 m)/(4.2 m/s) = 190 s. The
3.36.

current carries him downstream but doesn’t affect his motion in the perpendicular direction, from bank to bank.
IDENTIFY: Use the relation that relates the relative velocities.
G
SET UP: The relative velocities are the water relative to the earth, vW/E , the boat relative to the water,
G
G
G
G
vB/W , and the boat relative to the earth, v B/E . vB/E is due east, vW/E is due south and has magnitude
G
G
G
2.0 m/s. vB/W = 4.2 m/s. vB/E = vB/W + vW/E . The velocity addition diagram is given in Figure 3.36.

G
v
2.0 m/s

EXECUTE: (a) Find the direction of vB/W . sin θ = W/E =
. θ = 28.4°, north of east.
vB/W 4.2 m/s
2
2
− vW/E
= (4.2 m/s) 2 − (2.0 m/s) 2 = 3.7 m/s
(b) vB/E = vB/W

(c) t =

800 m 800 m
=
= 216 s.
vB/E
3.7 m/s

EVALUATE: It takes longer to cross the river in this problem than it did in Problem 3.35. In the direction
straight across the river (east) the component of his velocity relative to the earth is lass than 4.2 m/s.

Figure 3.36
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3-18
3.37.

Chapter 3
IDENTIFY: Relative velocity problem in two dimensions.

G
G
(a) SET UP: vP/A is the velocity of the plane relative to the air. The problem states that vP A has

magnitude 35 m/s and direction south.
G
G
vA/E is the velocity of the air relative to the earth. The problem states that vA/E is to the southwest
( 45° S of W) and has magnitude 10 m/s.
G
G
G
The relative velocity equation is vP/E = vP/A + vA/E .

Figure 3.37a
EXECUTE: (b) (vP/A ) x = 0, (vP/A ) y = −35 m/s

(vA/E ) x = −(10 m/s)cos 45° = −7.07 m/s,
(vA/E ) y = −(10 m/s)sin 45° = −7.07 m/s
(vP/E ) x = (vP/A ) x + (vA/E ) x = 0 − 7.07 m/s = −7.1 m/s
(vP/E ) y = (vP/A ) y + (vA/E ) y = −35 m/s − 7.07 m/s = −42 m/s
(c)

vP/E = (vP/E ) 2x + (vP/E ) 2y
vP/E = (−7.1 m/s) 2 + (−42 m/s) 2 = 43 m/s
tan φ =

(vP/E ) x −7.1
=
= 0.169

(vP/E ) y −42

φ = 9.6°; ( 9.6° west of south)
Figure 3.37b

3.38.

EVALUATE: The relative velocity addition diagram does not form a right triangle so the vector addition
must be done using components. The wind adds both southward and westward components to the velocity
of the plane relative to the ground.
IDENTIFY: Use the relation that relates the relative velocities.
G
SET UP: The relative velocities are the velocity of the plane relative to the ground, vP/G , the velocity of
G
G
G
the plane relative to the air, vP/A , and the velocity of the air relative to the ground, vA/G . vP/G must due
G
G
G
G
west and vA/G must be south. vA/G = 80 km/h and vP/A = 320 km/h. vP/G = vP/A + vA/G . The relative

velocity addition diagram is given in Figure 3.38.
v
80 km/h
EXECUTE: (a) sin θ = A/G =
and θ = 14°, north of west.
vP/A 320 km/h
2

2
(b) vP/G = vP/A
− vA/G
= (320 km/h)2 − (80.0 km/h)2 = 310 km/h.

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Motion in Two or Three Dimensions

3-19

EVALUATE: To travel due west the velocity of the plane relative to the air must have a westward
component and also a component that is northward, opposite to the wind direction.

Figure 3.38
3.39.

IDENTIFY: The resultant velocity, relative to the ground, is directly southward. This velocity is the sum of
the velocity of the bird relative to the air and the velocity of the air relative to the ground.
G
G
G
G
SET UP: vB/A = 100 km/h. vA/G = 40 km/h, east. vB/G = vB/A + vA/G .
G
EXECUTE: We want vB/G to be due south. The relative velocity addition diagram is shown in

Figure 3.39.

Figure 3.39
(a) sin φ =

vA/G
40 km/h
, φ = 24°, west of south.
=
vB/A 100 km/h

(b) vB/G = vB/A 2 − vA/G 2 = 91.7 km/h. t =

3.40.

d
500 km
=
= 5.5 h.
vB/G 91.7 km/h

EVALUATE: The speed of the bird relative to the ground is less than its speed relative to the air. Part of its
velocity relative to the air is directed to oppose the effect of the wind.
IDENTIFY: As the runner runs around the track, his speed stays the same but the direction of his velocity
changes so he has acceleration.
Δx
Δv
SET UP: (vx )av =
, (a x )av = x (and likewise for the y components). The coordinates of each point
Δt
Δt

are: A, (−50 m, 0); B, (0, + 50 m); C, (+50 m, 0); D, (0, − 50 m). At each point the velocity is tangent to

the circular path, as shown in Figure 3.40. The components (vx , v y ) of the velocity at each point are: A,
(0, + 6.0 m/s); B, (+6.0 m/s, 0); C, (0, − 6.0 m/s); D, ( − 6.0 m/s, 0).

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3-20

Chapter 3

Figure 3.40

2π r 2π (50 m)
=
= 52.4 s. A to B is one-quarter lap
v
6.0 m/s
Δx 0 − ( −50 m)
Δy +50 m − 0
and takes 14 (52.4 s) = 13.1 s. (vx )av =
=
= 3.8 m/s; (v y )av =
=
= 3.8 m/s.
Δt
13.1 s
Δt

13.1 s
Δv y 0 − 6.0 m/s
Δv
6.0 m/s − 0
=
= − 0.46 m/s 2
(a x )av = x =
= 0.46 m/s 2 ; (a y )av =
Δt
13.1 s
Δt
13.1 s
Δx +50 m − (−50 m)
Δy
(b) A to C: t = 12 (52.4 s) = 26.2 s. (vx )av =
=
= 3.8 m/s; (v y )av =
= 0.
Δt
Δt
26.2 s
Δv y −6.0 m/s − 6.0 m/s
Δv
=
= − 0.46 m/s 2 .
(a x )av = x = 0; (a y )av =
Δt
Δt
26.2 s
Δx 0 − 50 m

=
= − 3.8 m/s;
(c) C to D: t = 14 (52.4 s) = 13.1 s. (vx )av =
Δt
13.1 s
Δy −50 m − 0
Δv
−6.0 m/s − 0
(v y )av =
=
= − 3.8 m/s. (a x )av = x =
= − 0.46 m/s 2 ;
Δt
13.1 s
Δt
13.1 s
Δv y 0 − (−6.0 m/s)
(a y )av =
=
= 0.46 m/s 2 .
Δt
13.1 s
(d) A to A: Δx = Δy = 0 so (vx )av = (v y )av = 0, and Δvx = Δv y = 0 so (ax )av = (a y )av = 0.
EXECUTE: (a) A to B: The time for one full lap is t =

2
2
(e) For A to B: vav = (vx )av
+ (v y )av
= (3.8 m/s) 2 + (3.8 m/s) 2 = 5.4 m/s. The speed is constant so the

3.41.

average speed is 6.0 m/s. The average speed is larger than the magnitude of the average velocity because
the distance traveled is larger than the magnitude of the displacement.
(f) Velocity is a vector, with both magnitude and direction. The magnitude of the velocity is constant but
its direction is changing.
EVALUATE: For this motion the acceleration describes the rate of change of the direction of the velocity,
not the rate of change of the speed.
G
G
G dr
G dv
IDENTIFY: v =
and a =
dt
dt
d n
SET UP:
(t ) = nt n −1. At t = 1.00 s, a x = 4.00 m/s 2 and a y = 3.00 m/s 2 . At t = 0, x = 0 and
dt
y = 50.0 m.
dx
dv
= 2 Bt. a x = x = 2 B, which is independent of t. a x = 4.00 m/s 2 gives
dt
dt
dv y
dy
2

2
B = 2.00 m/s . v y =
= 3Dt . a y =
= 6 Dt. a y = 3.00 m/s 2 gives D = 0.500 m/s3. x = 0 at t = 0
dt
dt
gives A = 0. y = 50.0 m at t = 0 gives C = 50.0 m.
G
(b) At t = 0, vx = 0 and v y = 0, so v = 0. At t = 0, a x = 2 B = 4.00 m/s 2 and a y = 0, so
G
a = (4.00 m/s 2 )iˆ.
EXECUTE: (a) vx =

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Motion in Two or Three Dimensions

3-21

(c) At t = 10.0 s, vx = 2 (2.00 m/s 2 )(10.0 s) = 40.0 m/s and v y = 3(0.500 m/s3 )(10.0 s) 2 = 150 m/s.

v = vx2 + v 2y = 155 m/s.
(d) x = (2.00 m/s 2 )(10.0 s) 2 = 200 m, y = 50.0 m + (0.500 m/s3 )(10.0 s)3 = 550 m.
G
r = (200 m)iˆ + (550 m) ˆj.

3.42.

EVALUATE: The velocity and acceleration vectors as functions of time are
G
G
v (t ) = (2 Bt ) iˆ + (3Dt 2 ) ˆj and a (t ) = (2 B)iˆ + (6 Dt ) ˆj. The acceleration is not constant.
IDENTIFY: Use Eqs. (2.17) and (2.18).
SET UP: At the maximum height v y = 0.
EXECUTE: (a) vx = v0 x +

α

γ

α

β

γ

t 3 , v y = v0 y + β t − t 2 , and x = v0 xt + t 4 , y = v0 yt + t 2 − t 3.
3
2
12
2
6

γ

(b) Setting v y = 0 yields a quadratic in t , 0 = v0 y + β t − t 2 , which has as the positive solution
2
1⎡

t = β + β 2 + 2v0 yγ ⎤ = 13.59 s. Using this time in the expression for y(t) gives a maximum height of
⎥⎦
γ ⎢⎣

341 m.
(c) The path of the rocket is sketched in Figure 3.42.
(d) y = 0 gives 0 = v0 yt +

β

γ

γ

β

t 2 − t 3 and t 2 − t − v0 y = 0. The positive solution is t = 20.73 s. For this t,
2
6
6
2

x = 3.85 × 104 m.
EVALUATE: The graph in part (c) shows the path is not symmetric about the highest point and the time to
return to the ground is less than twice the time to the maximum height.

Figure 3.42
3.43.

3.44.

G
G
IDENTIFY: v = dr/dt. This vector will make a 45° angle with both axes when its x- and y-components
are equal.
d (t n )
SET UP:
= nt n −1.
dt
G
EXECUTE: v = 2btiˆ + 3ct 2 ˆj. vx = v y gives t = 2b 3c .
G
EVALUATE: Both components of v change with t.
IDENTIFY: Use the position vector of a dragonfly to determine information about its velocity vector and
acceleration vector.
SET UP: Use the definitions vx = dx/dt , v y = dy/dt , a x = dvx /dt , and a y = dv y /dt.

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3-22

Chapter 3
EXECUTE: (a) Taking derivatives of the position vector gives the components of the velocity vector:
vx (t ) = (0.180 m/s 2 )t , v y (t ) = ( −0.0450 m/s3 )t 2 . Use these components and the given direction:

tan 30.0o =

(0.0450 m/s 2 )t 2

, which gives t = 2.31 s.
(0.180 m/s 2 )t
(b) Taking derivatives of the velocity components gives the acceleration components:
a x = 0.180 m/s 2 , a y (t ) = −(0.0900 m/s3 )t. At t = 2.31 s, a x = 0.180 m/s 2 and a y = −0.208 m/s 2 , giving
0.208
, so θ = 49.1o clockwise from +x-axis.
0.180
EVALUATE: The acceleration is not constant, so we cannot use the standard kinematics formulas.
IDENTIFY: Given the velocity components of a plane, when will its velocity be perpendicular to its
acceleration?
SET UP: By definition, a x = dvx /dt , and a y = dv y /dt. When two vectors are perpendicular, their scalar
a = 0.275 m/s 2 . The direction is tan θ =

3.45.

product is zero.
G
EXECUTE: Taking the time derivative of the velocity vector gives a (t ) = (1.20 m/s 2 )iˆ + (−2.00 m/s 2 ) ˆj.
When the velocity and acceleration are perpendicular to each other,
G G
v ⋅ a = (1.20 m/s 2 ) 2 t + (12.0 m/s − (2.00 m/s 2 )t )(−2.00 m/s 2 ) = 0. Solving for t gives
(5.44 m 2 /s 4 )t = 24.0 m 2 /s3 , so t = 4.41 s.

3.46.

EVALUATE: There is only one instant at which the velocity and acceleration are perpendicular, so it is not
a general rule.
G
tG

dv
G G
IDENTIFY: r = r0 + ∫ v (t )dt and a = .
0
dt
SET UP: At t = 0, x0 = 0 and y0 = 0.

G
G ⎛
β ⎞ ⎛γ ⎞
EXECUTE: (a) Integrating, r = ⎜ α t − t 3 ⎟ iˆ + ⎜ t 2 ⎟ ˆj. Differentiating, a = (−2 β t ) iˆ + γ ˆj.
3
2
⎝
⎠ ⎝
⎠
(b) The positive time at which x = 0 is given by t 2 = 3α β . At this time, the y-coordinate is

y=
3.47.

γ
2

t2 =

3αγ 3(2.4 m/s)(4.0 m/s 2 )
=
= 9.0 m.
2β

2(1.6 m/s3 )

EVALUATE: The acceleration is not constant.
IDENTIFY: Once the rocket leaves the incline it moves in projectile motion. The acceleration along the
incline determines the initial velocity and initial position for the projectile motion.
SET UP: For motion along the incline let + x be directed up the incline. vx2 = v02x + 2a x ( x − x0 ) gives

vx = 2(1.25 m/s 2 )(200 m) = 22.36 m/s. When the projectile motion begins the rocket has v0 = 22.36 m/s
at 35.0° above the horizontal and is at a vertical height of (200.0 m) sin 35.0° = 114.7 m. For the
projectile motion let + x be horizontal to the right and let + y be upward. Let y = 0 at the ground. Then
y0 = 114.7 m, v0 x = v0 cos35.0° = 18.32 m/s, v0 y = v0 sin 35.0° = 12.83 m/s, a x = 0, a y = −9.80 m/s 2 . Let
x = 0 at point A, so x0 = (200.0 m)cos35.0° = 163.8 m.
EXECUTE: (a) At the maximum height v y = 0. v 2y = v02 y + 2a y ( y − y0 ) gives

y − y0 =

v 2y − v02 y

2a y

=

0 − (12.83 m/s) 2
2(−9.80 m/s 2 )

= 8.40 m and y = 114.7 m + 8.40 m = 123 m. The maximum height

above ground is 123 m.
(b) The time in the air can be calculated from the vertical component of the projectile motion:
y − y0 = − 114.7 m, v0 y = 12.83 m/s, a y = −9.80 m/s 2 . y − y0 = v0 yt + 12 a yt 2 gives

(4.90 m/s 2 )t 2 − (12.83 m/s)t − 114.7 m. The quadratic formula gives

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Motion in Two or Three Dimensions

t=

(

3-23

)

1
12.83 ± (12.83) 2 + 4(4.90)(114.7) s. The positive root is t = 6.32 s. Then
9.80

x − x0 = v0 xt + 12 a xt 2 = (18.32 m/s)(6.32 s) = 115.8 m and x = 163.8 m + 115.8 m = 280 m. The horizontal

3.48.

range of the rocket is 280 m.
EVALUATE: The expressions for h and R derived in Example 3.8 do not apply here. They are only for a
projectile fired on level ground.
IDENTIFY: The person moves in projectile motion. Use the results in Example 3.8 to determine how T, h
and D depend on g and set up a ratio.
2v sin α 0

v 2 sin 2 α 0
, the maximum height is h = 0
SET UP: From Example 3.8, the time in the air is t = 0
2g
g
and the horizontal range (called D in the problem) is D =

v02 sin 2α 0
. The person has the same v0 and α 0
g

on Mars as on the earth.

⎛g ⎞
⎛ gE ⎞
EXECUTE: tg = 2v0 sin α 0 , which is constant, so tE g E = tM g M . tM = ⎜ E ⎟ tE = ⎜
⎟ tE = 2.64tE .
⎝ gM ⎠
⎝ 0.379 g E ⎠
⎛g ⎞
v 2 sin 2 α 0
, which is constant, so hE g E = hM g M . hM = ⎜ E ⎟ hE = 2.64hE . Dg = v02 sin 2α 0 , which is
hg = 0
2
⎝ gM ⎠
⎛g ⎞
constant, so DE g E = DM g M . DM = ⎜ E ⎟ DE = 2.64 DE .
⎝ gM ⎠
EVALUATE: All three quantities are proportional to 1/g so all increase by the same factor of
g E / g M = 2.64.

3.49.

3.50.

IDENTIFY: The range for a projectile that lands at the same height from which it was launched is
v 2 sin 2α
.
R= 0
g
SET UP: The maximum range is for α = 45°.
EXECUTE: Assuming α = 45°, and R = 50 m, v0 = gR = 22 m/s.
EVALUATE: We have assumed that debris was launched at all angles, including the angle of 45° that
gives maximum range.
IDENTIFY: The velocity has a horizontal tangential component and a vertical component. The vertical
v2
component of acceleration is zero and the horizontal component is arad = x .
R
SET UP: Let + y be upward and + x be in the direction of the tangential velocity at the instant we are

considering.
EXECUTE: (a) The bird’s tangential velocity can be found from
circumference 2π (6.00 m)
vx =
=
= 7.54 m/s.
time of rotation
5.00 s
Thus its velocity consists of the components vx = 7.54 m/s and v y = 3.00 m/s. The speed relative to the
ground is then v = vx2 + v 2y = 8.11 m/s.
(b) The bird’s speed is constant, so its acceleration is strictly centripetal—entirely in the horizontal

v 2 (7.54 m/s) 2
= 9.48 m/s 2 .
direction, toward the center of its spiral path—and has magnitude arad = x =
6.00 m
r
3.00 m/s
= 21.7°.
(c) Using the vertical and horizontal velocity components θ = tan −1
7.54 m/s
EVALUATE: The angle between the bird’s velocity and the horizontal remains constant as the bird rises.

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3-24
3.51.

Chapter 3
IDENTIFY: Take + y to be downward. Both objects have the same vertical motion, with v0 y and

a y = + g . Use constant acceleration equations for the x and y components of the motion.
SET UP: Use the vertical motion to find the time in the air:

v0 y = 0, a y = 9.80 m/s 2 , y − y0 = 25 m, t = ?.
EXECUTE:

3.52.

y − y0 = v0 yt + 12 a yt 2 gives t = 2.259 s.

During this time the dart must travel 90 m, so the horizontal component of its velocity must be
x − x0 70 m
=
= 31 m/s.
v0 x =
t
2.25 s
EVALUATE: Both objects hit the ground at the same time. The dart hits the monkey for any muzzle
velocity greater than 31 m/s.
IDENTIFY: The person moves in projectile motion. Her vertical motion determines her time in the air.
SET UP: Take + y upward. v0 x = 15.0 m/s, v0 y = +10.0 m/s, a x = 0, a y = −9.80 m/s 2 .
EXECUTE: (a) Use the vertical motion to find the time in the air: y − y0 = v0 yt + 12 a yt 2 with

y − y0 = −30.0 m gives −30.0 m = (10.0 m/s)t − (4.90 m/s 2 )t 2 . The quadratic formula gives
t=

(

)

1
+10.0 ± (−10.0) 2 − 4(4.9)( −30) s. The positive solution is t = 3.70 s. During this time she
2(4.9)

travels a horizontal distance x − x0 = v0 xt + 12 a xt 2 = (15.0 m/s)(3.70 s) = 55.5 m. She will land 55.5 m south
of the point where she drops from the helicopter and this is where the mats should have been placed.
(b) The x-t, y-t, vx -t and v y -t graphs are sketched in Figure 3.52.
EVALUATE: If she had dropped from rest at a height of 30.0 m it would have taken her
2(30.0 m)

t=
= 2.47 s. She is in the air longer than this because she has an initial vertical component of
9.80 m/s 2
velocity that is upward.

Figure 3.52
3.53.

IDENTIFY: The cannister moves in projectile motion. Its initial velocity is horizontal. Apply constant
acceleration equations for the x and y components of motion.
SET UP:

Take the origin of coordinates at the point
where the canister is released. Take +y to be
upward. The initial velocity of the canister is
the velocity of the plane, 64.0 m/s in the
+x-direction.

Figure 3.53

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Motion in Two or Three Dimensions

3-25

Use the vertical motion to find the time of fall:
t = ?, v0 y = 0, a y = −9.80 m/s 2 , y − y0 = −90.0 m (When the canister reaches the ground it is 90.0 m

below the origin.)
y − y0 = v0 yt + 12 a yt 2
EXECUTE: Since v0 y = 0, t =

2( y − y0 )
2(−90.0 m)
=
= 4.286 s.
ay
−9.80 m/s 2

SET UP: Then use the horizontal component of the motion to calculate how far the canister falls in this
time:
x − x0 = ?, a x − 0, v0 x = 64.0 m/s
EXECUTE:

3.54.

x − x0 = v0t + 12 at 2 = (64.0 m/s)(4.286 s) + 0 = 274 m.

EVALUATE: The time it takes the cannister to fall 90.0 m, starting from rest, is the time it travels
horizontally at constant speed.
IDENTIFY: The shell moves as a projectile. To just clear the top of the cliff, the shell must have
y − y0 = 25.0 m when it has x − x0 = 60.0 m.
SET UP: Let + y be upward. a x = 0, a y = − g . v0 x = v0 cos 43°, v0 y = v0 sin 43°.
EXECUTE: (a) horizontal motion: x − x0 = v0 xt so t =

60.0 m
.
(v0 cos 43°)

vertical motion: y − y0 = v0 yt + 12 a yt 2 gives 25.0m = (v0 sin 43.0°) t + 12 ( −9.80m/s 2 ) t 2 .
Solving these two simultaneous equations for v0 and t gives v0 = 32.6 m/s and t = 2.51 s.
(b) v y when shell reaches cliff:

v y = v0 y + a yt = (32.6 m/s) sin 43.0° − (9.80 m/s 2 )(2.51 s) = −2.4 m/s

The shell is traveling downward when it reaches the cliff, so it lands right at the edge of the cliff.
v0 y
EVALUATE: The shell reaches its maximum height at t = −
= 2.27 s, which confirms that at
ay
3.55.

t = 2.51 s it has passed its maximum height and is on its way down when it strikes the edge of the cliff.
IDENTIFY: The suitcase moves in projectile motion. The initial velocity of the suitcase equals the velocity
of the airplane.
SET UP: Take + y to be upward. a x = 0, a y = − g .
EXECUTE: Use the vertical motion to find the time it takes the suitcase to reach the ground:
v0 y = v0 sin23°, a y = −9.80 m/s 2 , y − y0 = −114 m, t = ? y − y0 = v0 yt + 12 a yt 2 gives t = 9.60 s.

The distance the suitcase travels horizontally is x − x0 = v0 x = (v0 cos23.0°)t = 795 m.
EVALUATE: An object released from rest at a height of 114 m strikes the ground at
2( y − y0 )
t=
= 4.82 s. The suitcase is in the air much longer than this since it initially has an upward
−g
3.56.

component of velocity.

IDENTIFY: The equipment moves in projectile motion. The distance D is the horizontal range of the
equipment plus the distance the ship moves while the equipment is in the air.
SET UP: For the motion of the equipment take + x to be to the right and + y to be upward. Then a x = 0,
a y = −9.80 m/s 2 , v0 x = v0 cos α 0 = 7.50 m/s and v0 y = v0 sin α 0 = 13.0 m/s. When the equipment lands in

the front of the ship, y − y0 = −8.75 m.
EXECUTE: Use the vertical motion of the equipment to find its time in the air: y − y0 = v0 yt + 12 a yt 2 gives
t=

(

)

1
13.0 ± (−13.0) 2 + 4(4.90)(8.75) s. The positive root is t = 3.21 s. The horizontal range of the
9.80

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